Q: Q1. Which of the following examples represent (a) periodic motion, (b) simple harmonic motion, and (c) non-periodic motion?\n(i) A swimmer completing one trip from one bank of a river

Let's classify each motion based on the definitions:\n\n* **Periodic Motion:** A motion that repeats itself after a fixed interval of time.\n* **Simple Harmonic Motion (SHM):** A special type of periodic motion where the restoring force (or torque) is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium. It can be represented by a sine or cosine function.\n* **Non-periodic Motion:** A motion that does not repeat itself at regular intervals.\n\nApplying these definitions to the given examples:\n\n(i) **A swimmer completing one trip from one bank of a river to the other and back.**\n This is generally **non-periodic motion**. A swimmer's speed and path might vary with fatigue, river currents, or other factors, making it unlikely for the motion to repeat in exactly the same time interval or manner. Even if we idealize it, it's not a natural oscillation or rotation.\n\n(ii) **A freely suspended bar magnet displaced from its N-S direction and released.**\n This is an example of **simple harmonic motion (SHM)** (for small angular displacements). When displaced, a restoring torque acts on the magnet, tending to bring it back to its equilibrium (N-S) position. For small angles, this torque is approximately proportional to the angular displacement, leading to SHM.\n\n(iii) **A hydrogen molecule rotating about its centre of mass.**\n This is an example of **periodic motion**. The molecule rotates continuously, repeating its orientation after every full rotation. However, it is not SHM because there is no restoring force/torque proportional to angular displacement that would bring it back to a specific equilibrium orientation (assuming free rotation in space).\n\n(iv) **An arrow released from a bow.**\n This is an example of **non-periodic motion**. Once released, the arrow follows a projectile path and does not repeat its motion. It travels from one point to another without returning or oscillating.

Q: Q5. A body oscillates with SHM according to the equation (in SI units), x = 5 cos (2πt + π/4). At t = 1.5 s, calculate the (a) displacement, (b) speed, (c) acceleration.

Given the equation for SHM: x = 5 cos (2πt + π/4).\nComparing this with the standard SHM equation x = A cos (ωt + φ), we have:\nAmplitude A = 5 m\nAngular frequency ω = 2π rad/s\nInitial phase φ = π/4 rad\n\nAt t = 1.5 s:\n\n(a) Displacement (x):\nx = 5 cos (2π * 1.5 + π/4)\nx = 5 cos (3π + π/4)\nx = 5 cos (12π/4 + π/4)\nx = 5 cos (13π/4)\nWe know that cos(θ + 2nπ) = cos(θ). So, cos(13π/4) = cos(3π + π/4) = cos(π + π/4) = -cos(π/4).\nx = 5 * (-cos(π/4))\nx = 5 * (-1/√2)\nx = -5/√2 m ≈ -3.536 m\n\n(b) Speed (v):\nThe velocity equation is v = dx/dt.\nv = d/dt [5 cos (2πt + π/4)]\nv = 5 * [-sin (2πt + π/4)] * (2π)\nv = -10π sin (2πt + π/4)\nAt t = 1.5 s:\nv = -10π sin (2π * 1.5 + π/4)\nv = -10π sin (3π + π/4)\nv = -10π sin (13π/4)\nWe know that sin(θ + 2nπ) = sin(θ). So, sin(13π/4) = sin(3π + π/4) = sin(π + π/4) = -sin(π/4).\nv = -10π * (-sin(π/4))\nv = -10π * (-1/√2)\nv = 10π/√2 m/s ≈ 22.21 m/s\n\n(c) Acceleration (a):\nThe acceleration equation is a = dv/dt.\na = d/dt [-10π sin (2πt + π/4)]\na = -10π * [cos (2πt + π/4)] * (2π)\na = -20π² cos (2πt + π/4)\nAt t = 1.5 s:\na = -20π² cos (2π * 1.5 + π/4)\na = -20π² cos (3π + π/4)\na = -20π² cos (13π/4)\nAs calculated for displacement, cos(13π/4) = -cos(π/4) = -1/√2.\na = -20π² * (-1/√2)\na = 20π²/√2 m/s² ≈ 139.38 m/s²

Q: Q7. A particle executes SHM with a period of 0.05 s and an amplitude of 2 cm. Find (a) the maximum velocity, (b) the maximum acceleration, (c) the velocity and acceleration at a displa

Given:\nPeriod, T = 0.05 s\nAmplitude, A = 2 cm = 0.02 m\n\nFirst, calculate the angular frequency (ω):\nω = 2π / T\nω = 2π / 0.05\nω = 40π rad/s\n\n(a) Maximum velocity (v_max):\nThe maximum velocity in SHM is given by v_max = Aω.\nv_max = (0.02 m) * (40π rad/s)\nv_max = 0.8π m/s\nv_max ≈ 0.8 * 3.14159 m/s\nv_max ≈ 2.513 m/s\n\n(b) Maximum acceleration (a_max):\nThe maximum acceleration in SHM is given by a_max = Aω².\na_max = (0.02 m) * (40π rad/s)²\na_max = 0.02 * (1600π²) m/s²\na_max = 32π² m/s²\na_max ≈ 32 * (3.14159)² m/s²\na_max ≈ 32 * 9.8696 m/s²\na_max ≈ 315.83 m/s²\n\n(c) Velocity (v) and acceleration (a) at a displacement of 1 cm:\nGiven displacement, x = 1 cm = 0.01 m\n\nVelocity at displacement x:\nThe velocity in SHM is given by v = ±ω√(A² - x²).\nv = ±(40π)√((0.02)² - (0.01)²)\nv = ±(40π)√(0.0004 - 0.0001)\nv = ±(40π)√(0.0003)\nv = ±(40π) * 0.01√3\nv = ±0.4π√3 m/s\nv ≈ ±0.4 * 3.14159 * 1.732 m/s\nv ≈ ±2.177 m/s\n\nAcceleration at displacement x:\nThe acceleration in SHM is given by a = -ω²x.\na = -(40π)² * (0.01)\na = -(1600π²) * 0.01\na = -16π² m/s²\na ≈ -16 * (3.14159)² m/s²\na ≈ -16 * 9.8696 m/s²\na ≈ -157.91 m/s²

Q: Q8. A body of mass 10 kg is attached to a spring of spring constant 250 N m⁻¹. The body is displaced by 0.1 m from its equilibrium position and released. Find (a) the angular frequency

Given:\nMass of the body, m = 10 kg\nSpring constant, k = 250 N m⁻¹\nInitial displacement (Amplitude), A = 0.1 m\n\n(a) Angular frequency (ω):\nFor a spring-mass system, the angular frequency is given by:\nω = √(k/m)\nω = √(250 N m⁻¹ / 10 kg)\nω = √(25) rad/s\nω = 5 rad/s\n\n(b) Period of oscillation (T):\nThe period of oscillation is related to the angular frequency by:\nT = 2π / ω\nT = 2π / 5 s\nT ≈ 1.257 s\n\n(c) Maximum speed (v_max):\nThe maximum speed in SHM is given by:\nv_max = Aω\nv_max = (0.1 m) * (5 rad/s)\nv_max = 0.5 m/s\n\n(d) Maximum acceleration (a_max):\nThe maximum acceleration in SHM is given by:\na_max = Aω²\na_max = (0.1 m) * (5 rad/s)²\na_max = 0.1 * 25 m/s²\na_max = 2.5 m/s²\n\n(e) Total energy (E):\nThe total energy of a body in SHM is constant and can be calculated using the formula:\nE = (1/2)kA²\nE = (1/2) * (250 N m⁻¹) * (0.1 m)²\nE = (1/2) * 250 * 0.01 J\nE = 125 * 0.01 J\nE = 1.25 J

Q: Q13. A body describes simple harmonic motion with amplitude 5 cm and period 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.

Given:\nAmplitude, A = 5 cm = 0.05 m\nPeriod, T = 0.2 s\n\nFirst, calculate the angular frequency (ω):\nω = 2π/T = 2π/0.2 = 10π rad/s\n\nNow, we calculate velocity (v) and acceleration (a) for the given displacements:\nVelocity, v = ω√(A² - x²)\nAcceleration, a = -ω²x\n\n(a) When displacement x = 5 cm = 0.05 m (at the extreme position):\nVelocity, v = 10π√((0.05)² - (0.05)²) = 10π√(0) = 0 m/s\nAcceleration, a = -(10π)²(0.05) = -100π²(0.05) = -5π² m/s²\n\n(b) When displacement x = 3 cm = 0.03 m:\nVelocity, v = 10π√((0.05)² - (0.03)²) = 10π√(0.0025 - 0.0009) = 10π√(0.0016) = 10π(0.04) = 0.4π m/s\nAcceleration, a = -(10π)²(0.03) = -100π²(0.03) = -3π² m/s²\n\n(c) When displacement x = 0 cm = 0 m (at the mean position):\nVelocity, v = 10π√((0.05)² - (0)²) = 10π(0.05) = 0.5π m/s\nAcceleration, a = -(10π)²(0) = 0 m/s²\n\nSummary of results:\n(a) At x = 5 cm: v = 0 m/s, a = -5π² m/s²\n(b) At x = 3 cm: v = 0.4π m/s, a = -3π² m/s²\n(c) At x = 0 cm: v = 0.5π m/s, a = 0 m/s²

Question 1

Q1. Which of the following examples represent (a) periodic motion, (b) simple harmonic motion, and (c) non-periodic motion?
(i) A swimmer completing one trip from one bank of a river

Accepted Answer

Let's classify each motion based on the definitions:

*   **Periodic Motion:** A motion that repeats itself after a fixed interval of time.
*   **Simple Harmonic Motion (SHM):** A special type of periodic motion where the restoring force (or torque) is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium. It can be represented by a sine or cosine function.
*   **Non-periodic Motion:** A motion that does not repeat itself at regular intervals.

Applying these definitions to the given examples:

(i) **A swimmer completing one trip from one bank of a river to the other and back.**
    This is generally **non-periodic motion**. A swimmer's speed and path might vary with fatigue, river currents, or other factors, making it unlikely for the motion to repeat in exactly the same time interval or manner. Even if we idealize it, it's not a natural oscillation or rotation.

(ii) **A freely suspended bar magnet displaced from its N-S direction and released.**
    This is an example of **simple harmonic motion (SHM)** (for small angular displacements). When displaced, a restoring torque acts on the magnet, tending to bring it back to its equilibrium (N-S) position. For small angles, this torque is approximately proportional to the angular displacement, leading to SHM.

(iii) **A hydrogen molecule rotating about its centre of mass.**
    This is an example of **periodic motion**. The molecule rotates continuously, repeating its orientation after every full rotation. However, it is not SHM because there is no restoring force/torque proportional to angular displacement that would bring it back to a specific equilibrium orientation (assuming free rotation in space).

(iv) **An arrow released from a bow.**
    This is an example of **non-periodic motion**. Once released, the arrow follows a projectile path and does not repeat its motion. It travels from one point to another without returning or oscillating.

Question 2

Q2. Which of the following functions of time represent (a) simple harmonic motion, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of perio

Accepted Answer

To classify these functions, we recall the definitions:

*   **Simple Harmonic Motion (SHM):** A motion described by a single sine or cosine function of time, i.e., of the form `A sin(ω't + φ)` or `A cos(ω't + φ)`. The second derivative with respect to time must be proportional to the negative of the displacement (`d²x/dt² = -k x`).
*   **Periodic Motion:** A motion `x(t)` is periodic if `x(t) = x(t + T)` for some constant period `T`. The period for `sin(ω't)` or `cos(ω't)` is `T = 2π/ω'`.
*   **Non-periodic Motion:** A motion that does not repeat itself at regular intervals.

Let's analyze each function:

(a) **sin ωt – cos ωt**
    This can be rewritten using trigonometric identities: `R sin(ωt + α) = R sin(ωt)cos(α) + R cos(ωt)sin(α)`. Comparing with `sin ωt – cos ωt`, we have `R cos(α) = 1` and `R sin(α) = -1`. Squaring and adding gives `R²(cos²α + sin²α) = 1² + (-1)² = 2`, so `R = √2`. Dividing gives `tan(α) = -1`, so `α = -π/4` (or `7π/4`).
    Thus, `sin ωt – cos ωt = √2 sin(ωt - π/4)`. This is of the form `A sin(ω't + φ)`.
    **Classification:** Simple Harmonic Motion (SHM)
    **Period:** `T = 2π/ω`

(b) **sin² ωt**
    Using the trigonometric identity `sin²θ = (1 - cos 2θ)/2`:
    `sin² ωt = (1 - cos 2ωt)/2 = 1/2 - (1/2)cos(2ωt)`. This function is periodic. It oscillates about `1/2` with an angular frequency of `2ω`. Since it contains a constant term `1/2`, it is not SHM about the origin (x=0). SHM is typically defined as oscillation about a mean position, which is usually taken as zero. If the mean position is shifted, it can be considered SHM. However, in the strict sense of `d²x/dt² = -k x`, it's not SHM.
    **Classification:** Periodic but not Simple Harmonic Motion (due to the constant term and oscillation not being about zero).
    **Period:** The angular frequency is `2ω`, so `T = 2π/(2ω) = π/ω`.

(c) **3 cos (π/4 – 2ωt)**
    This is of the form `A cos(ω't + φ)`, where `A = 3`, `ω' = 2ω`, and `φ = π/4`. (Note: `cos(A-B) = cos(B-A)`, so `cos(π/4 - 2ωt) = cos(2ωt - π/4)`).
    **Classification:** Simple Harmonic Motion (SHM)
    **Period:** The angular frequency is `2ω`, so `T = 2π/(2ω) = π/ω`.

(d) **cos ωt + cos 3ωt + cos 5ωt**
    This is a superposition of three simple harmonic motions with different angular frequencies (ω, 3ω, 5ω). The resultant motion will be periodic if the ratio of the angular frequencies is rational. Here, the ratios are rational (1:3:5). The fundamental period will be the least common multiple (LCM) of the individual periods (`2π/ω`, `2π/3ω`, `2π/5ω`). The LCM is `2π/ω`.
    However, a superposition of multiple SHMs with different frequencies is generally not SHM itself, as it cannot be expressed as a single sine or cosine function.
    **Classification:** Periodic but not Simple Harmonic Motion.
    **Period:** `T = 2π/ω` (the fundamental period).

(e) **exp (–ω²t²)**
    This is an exponential decay function. As `t` increases, `e^(-ω²t²)` approaches 0. It does not repeat its values.
    **Classification:** Non-periodic Motion.
    **Period:** Not applicable.

(f) **1 + ωt + ω²t²**
    This is a polynomial function of `t`. As `t` increases, the value of the function continuously increases (for positive `ω`). It does not repeat its values.
    **Classification:** Non-periodic Motion.
    **Period:** Not applicable.

Question 3

Q3. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and choose the midpoint O as the o

Accepted Answer

The problem describes a particle executing linear simple harmonic motion (SHM). We are given the total distance between the extreme points, the speed at a certain displacement, and we need to find the time period.

Question 4

Q4. The displacement of a particle executing simple harmonic motion is given by y = A sin (ωt + φ). If the displacement of the particle at t = 0 is 2 cm and its velocity is 4 cm/s, fin

Accepted Answer

We are given the general equation for displacement in SHM and initial conditions to find the amplitude and initial phase angle.

Question 5

Q5. A body oscillates with SHM according to the equation (in SI units), x = 5 cos (2πt + π/4). At t = 1.5 s, calculate the (a) displacement, (b) speed, (c) acceleration.

Accepted Answer

Given the equation for SHM: x = 5 cos (2πt + π/4).
Comparing this with the standard SHM equation x = A cos (ωt + φ), we have:
Amplitude A = 5 m
Angular frequency ω = 2π rad/s
Initial phase φ = π/4 rad

At t = 1.5 s:

(a) Displacement (x):
x = 5 cos (2π * 1.5 + π/4)
x = 5 cos (3π + π/4)
x = 5 cos (12π/4 + π/4)
x = 5 cos (13π/4)
We know that cos(θ + 2nπ) = cos(θ). So, cos(13π/4) = cos(3π + π/4) = cos(π + π/4) = -cos(π/4).
x = 5 * (-cos(π/4))
x = 5 * (-1/√2)
x = -5/√2 m ≈ -3.536 m

(b) Speed (v):
The velocity equation is v = dx/dt.
v = d/dt [5 cos (2πt + π/4)]
v = 5 * [-sin (2πt + π/4)] * (2π)
v = -10π sin (2πt + π/4)
At t = 1.5 s:
v = -10π sin (2π * 1.5 + π/4)
v = -10π sin (3π + π/4)
v = -10π sin (13π/4)
We know that sin(θ + 2nπ) = sin(θ). So, sin(13π/4) = sin(3π + π/4) = sin(π + π/4) = -sin(π/4).
v = -10π * (-sin(π/4))
v = -10π * (-1/√2)
v = 10π/√2 m/s ≈ 22.21 m/s

(c) Acceleration (a):
The acceleration equation is a = dv/dt.
a = d/dt [-10π sin (2πt + π/4)]
a = -10π * [cos (2πt + π/4)] * (2π)
a = -20π² cos (2πt + π/4)
At t = 1.5 s:
a = -20π² cos (2π * 1.5 + π/4)
a = -20π² cos (3π + π/4)
a = -20π² cos (13π/4)
As calculated for displacement, cos(13π/4) = -cos(π/4) = -1/√2.
a = -20π² * (-1/√2)
a = 20π²/√2 m/s² ≈ 139.38 m/s²

Question 6

Q6. The corresponding reference circle for a particle in SHM is shown in Fig. 14.21. The particle is at P at t = 0. The radius of the circle is A, the angular speed of the projection o

Accepted Answer

### Understanding the Reference Circle for SHM

Simple Harmonic Motion (SHM) can be visualized as the projection of uniform circular motion onto a diameter of the circle. This circle is called the reference circle. Let's consider a particle undergoing uniform circular motion with angular speed ω and radius A. If we project the position of this particle onto the x-axis (or y-axis), the projected point will execute SHM.

**Given:**
*   Radius of the reference circle = A
*   Angular speed of the particle = ω
*   At t = 0, the particle is at point P. Let's assume P makes an angle φ with the positive x-axis at t=0 (initial phase).

#### 1. Period of Oscillation (T)

The angular speed ω is the rate of change of angular displacement. For one complete revolution (2π radians), the time taken is the period T.

ω = Angular displacement / Time taken
ω = 2π / T

Therefore, the period of oscillation T is:
**T = 2π / ω**

#### 2. Amplitude (A)

The amplitude of SHM is the maximum displacement from the equilibrium position. In the reference circle, the maximum displacement of the projection from the center (equilibrium position) is equal to the radius of the circle.

Therefore, the amplitude A is:
**A = Radius of the reference circle**

#### 3. Displacement (x) at any time t

Let the particle start at an initial angular position φ (at t=0) with respect to the positive x-axis. After time t, the particle will have rotated by an angle ωt. So, its total angular position at time t will be (ωt + φ).

The displacement x of the projection on the x-axis is given by the x-coordinate of the particle on the reference circle:

**x(t) = A cos (ωt + φ)**

If the particle starts at P on the positive x-axis at t=0 (i.e., φ=0), then x(t) = A cos(ωt).
If the particle starts at P on the positive y-axis at t=0 (i.e., φ=π/2), then x(t) = A cos(ωt + π/2) = -A sin(ωt).

#### 4. Velocity (v) at any time t

The velocity of the particle in SHM is the time derivative of its displacement.

v(t) = dx/dt
v(t) = d/dt [A cos (ωt + φ)]
v(t) = -Aω sin (ωt + φ)

Alternatively, we can think of the velocity vector of the particle in uniform circular motion, which is tangential to the circle and has magnitude v_circ = Aω. The projection of this tangential velocity vector onto the x-axis gives the velocity of the SHM.

**v(t) = -Aω sin (ωt + φ)**

#### 5. Acceleration (a) at any time t

The acceleration of the particle in SHM is the time derivative of its velocity.

a(t) = dv/dt
a(t) = d/dt [-Aω sin (ωt + φ)]
a(t) = -Aω * [ω cos (ωt + φ)]
a(t) = -Aω² cos (ωt + φ)

Since x(t) = A cos (ωt + φ), we can write acceleration in terms of displacement:

**a(t) = -ω² x(t)**

This shows that the acceleration in SHM is directly proportional to the displacement and is always directed towards the equilibrium position (opposite to the displacement). This is the defining characteristic of SHM.

**Summary:**
*   **Period (T):** T = 2π / ω
*   **Amplitude (A):** A = Radius of the reference circle
*   **Displacement (x):** x(t) = A cos (ωt + φ)
*   **Velocity (v):** v(t) = -Aω sin (ωt + φ)
*   **Acceleration (a):** a(t) = -Aω² cos (ωt + φ) = -ω² x(t)

Question 7

Q7. A particle executes SHM with a period of 0.05 s and an amplitude of 2 cm. Find (a) the maximum velocity, (b) the maximum acceleration, (c) the velocity and acceleration at a displa

Accepted Answer

Given:
Period, T = 0.05 s
Amplitude, A = 2 cm = 0.02 m

First, calculate the angular frequency (ω):
ω = 2π / T
ω = 2π / 0.05
ω = 40π rad/s

(a) Maximum velocity (v_max):
The maximum velocity in SHM is given by v_max = Aω.
v_max = (0.02 m) * (40π rad/s)
v_max = 0.8π m/s
v_max ≈ 0.8 * 3.14159 m/s
v_max ≈ 2.513 m/s

(b) Maximum acceleration (a_max):
The maximum acceleration in SHM is given by a_max = Aω².
a_max = (0.02 m) * (40π rad/s)²
a_max = 0.02 * (1600π²) m/s²
a_max = 32π² m/s²
a_max ≈ 32 * (3.14159)² m/s²
a_max ≈ 32 * 9.8696 m/s²
a_max ≈ 315.83 m/s²

(c) Velocity (v) and acceleration (a) at a displacement of 1 cm:
Given displacement, x = 1 cm = 0.01 m

Velocity at displacement x:
The velocity in SHM is given by v = ±ω√(A² - x²).
v = ±(40π)√((0.02)² - (0.01)²)
v = ±(40π)√(0.0004 - 0.0001)
v = ±(40π)√(0.0003)
v = ±(40π) * 0.01√3
v = ±0.4π√3 m/s
v ≈ ±0.4 * 3.14159 * 1.732 m/s
v ≈ ±2.177 m/s

Acceleration at displacement x:
The acceleration in SHM is given by a = -ω²x.
a = -(40π)² * (0.01)
a = -(1600π²) * 0.01
a = -16π² m/s²
a ≈ -16 * (3.14159)² m/s²
a ≈ -16 * 9.8696 m/s²
a ≈ -157.91 m/s²

Question 8

Q8. A body of mass 10 kg is attached to a spring of spring constant 250 N m⁻¹. The body is displaced by 0.1 m from its equilibrium position and released. Find (a) the angular frequency

Accepted Answer

Given:
Mass of the body, m = 10 kg
Spring constant, k = 250 N m⁻¹
Initial displacement (Amplitude), A = 0.1 m

(a) Angular frequency (ω):
For a spring-mass system, the angular frequency is given by:
ω = √(k/m)
ω = √(250 N m⁻¹ / 10 kg)
ω = √(25) rad/s
ω = 5 rad/s

(b) Period of oscillation (T):
The period of oscillation is related to the angular frequency by:
T = 2π / ω
T = 2π / 5 s
T ≈ 1.257 s

(c) Maximum speed (v_max):
The maximum speed in SHM is given by:
v_max = Aω
v_max = (0.1 m) * (5 rad/s)
v_max = 0.5 m/s

(d) Maximum acceleration (a_max):
The maximum acceleration in SHM is given by:
a_max = Aω²
a_max = (0.1 m) * (5 rad/s)²
a_max = 0.1 * 25 m/s²
a_max = 2.5 m/s²

(e) Total energy (E):
The total energy of a body in SHM is constant and can be calculated using the formula:
E = (1/2)kA²
E = (1/2) * (250 N m⁻¹) * (0.1 m)²
E = (1/2) * 250 * 0.01 J
E = 125 * 0.01 J
E = 1.25 J

Question 9

Q9. A simple pendulum of length $l$ and having a bob of mass $M$ is suspended in a car. The car is accelerating horizontally with an acceleration $a$. Find the period of oscillation of

Accepted Answer

When a simple pendulum is suspended in a car accelerating horizontally with acceleration $a$, the bob experiences two forces in the non-inertial frame of the car: its weight $Mg$ acting vertically downwards, and a pseudo force $Ma$ acting horizontally opposite to the direction of acceleration. The effective gravitational acceleration ($g_{eff}$) is the resultant of the actual gravitational acceleration $g$ and the acceleration $a$. The period of oscillation of the pendulum will then be determined by this effective gravitational acceleration.

Let the acceleration of the car be $a$ in the positive x-direction. The forces acting on the bob are:
1.  Gravitational force: $Mg$ (downwards)
2.  Pseudo force: $Ma$ (horizontally opposite to $a$, i.e., in the negative x-direction)

The resultant force on the bob, $F_{net}$, in the frame of the car is the vector sum of these two forces. The effective acceleration experienced by the bob, $g_{eff}$, is given by:
$g_{eff} = \sqrt{g^2 + a^2}$

For a simple pendulum, the period of oscillation ($T$) is given by the formula:
$T = 2\pi \sqrt{\frac{l}{g_{eff}}}$

Substituting the expression for $g_{eff}$:
$T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + a^2}}}$

This is the period of oscillation of the pendulum in the accelerating car.

Question 10

Q10. A body oscillates with SHM according to the equation $x = A \cos(\omega t + \phi)$. At $t = 0$, the displacement is $x_0$ and the velocity is $v_0$. Show that
(a) $	an \phi = -v_

Accepted Answer

Given the equation for displacement in SHM: $x(t) = A \cos(\omega t + \phi)$

First, let's find the velocity $v(t)$ by differentiating the displacement equation with respect to time:
$v(t) = \frac{dx}{dt} = \frac{d}{dt} [A \cos(\omega t + \phi)]$
$v(t) = -A \omega \sin(\omega t + \phi)$

Now, we apply the initial conditions at $t = 0$:
At $t = 0$, $x = x_0$ and $v = v_0$.

Substitute $t = 0$ into the displacement equation:
$x_0 = A \cos(\omega \cdot 0 + \phi)$
$x_0 = A \cos(\phi)$  --- (1)

Substitute $t = 0$ into the velocity equation:
$v_0 = -A \omega \sin(\omega \cdot 0 + \phi)$
$v_0 = -A \omega \sin(\phi)$ --- (2)

**(a) To show $	an \phi = -v_0 / (\omega x_0)$:**
Divide equation (2) by equation (1):
$\frac{v_0}{x_0} = \frac{-A \omega \sin(\phi)}{A \cos(\phi)}$
$\frac{v_0}{x_0} = -\omega \frac{\sin(\phi)}{\cos(\phi)}$
$\frac{v_0}{x_0} = -\omega 	an(\phi)$

Rearranging to solve for $	an(\phi)$:
$	an(\phi) = -\frac{v_0}{\omega x_0}$
Hence, part (a) is shown.

**(b) To show $A = \sqrt{x_0^2 + (v_0 / \omega)^2}$:**
From equation (1), we have $x_0 = A \cos(\phi)$, so $\cos(\phi) = \frac{x_0}{A}$.
From equation (2), we have $v_0 = -A \omega \sin(\phi)$, so $\sin(\phi) = -\frac{v_0}{A \omega}$.

We know the trigonometric identity: $\sin^2(\phi) + \cos^2(\phi) = 1$.
Substitute the expressions for $\sin(\phi)$ and $\cos(\phi)$:
$\left(-\frac{v_0}{A \omega}ight)^2 + \left(\frac{x_0}{A}ight)^2 = 1$
$\frac{v_0^2}{A^2 \omega^2} + \frac{x_0^2}{A^2} = 1$

Multiply the entire equation by $A^2$:
$\frac{v_0^2}{\omega^2} + x_0^2 = A^2$

Rearranging to solve for $A$:
$A^2 = x_0^2 + \left(\frac{v_0}{\omega}ight)^2$
$A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}ight)^2}$
Hence, part (b) is shown.

Question 11

Q11. A mass attached to a spring is free to oscillate, with its displacement given by $x(t) = A \cos(\omega t + \phi)$.
(a) What is the phase at $t = 0$?
(b) What is the phase at $t =

Accepted Answer

The general equation for displacement in SHM is $x(t) = A \cos(\omega t + \phi)$. The term $(\omega t + \phi)$ is called the phase of the oscillation.

**(a) Phase at $t = 0$:**
Substitute $t = 0$ into the phase expression:
Phase $= (\omega \cdot 0 + \phi) = \phi$
So, the phase at $t = 0$ is $\phi$.

**(b) Phase at $t = T/2$ where $T$ is the period of oscillation:**
We know that the angular frequency $\omega$ is related to the period $T$ by $\omega = \frac{2\pi}{T}$.
Substitute $t = T/2$ into the phase expression:
Phase $= (\omega \cdot \frac{T}{2} + \phi)$
Phase $= (\frac{2\pi}{T} \cdot \frac{T}{2} + \phi)$
Phase $= (\pi + \phi)$
So, the phase at $t = T/2$ is $(\pi + \phi)$.

**(c) Phase difference between $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \cos(\omega t + \phi_2)$:**
To find the phase difference, we need to express both displacements using the same trigonometric function (either both sine or both cosine).
We know that $\sin(	heta) = \cos(	heta - \frac{\pi}{2})$ or $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$.
Let's convert $x_1$ to a cosine function:
$x_1 = A \sin(\omega t + \phi_1) = A \cos((\omega t + \phi_1) - \frac{\pi}{2})$
$x_1 = A \cos(\omega t + \phi_1 - \frac{\pi}{2})$

Now we have:
Phase of $x_1$: $\Phi_1 = \omega t + \phi_1 - \frac{\pi}{2}$
Phase of $x_2$: $\Phi_2 = \omega t + \phi_2$

The phase difference $\Delta \Phi$ is the difference between these two phases:
$\Delta \Phi = \Phi_2 - \Phi_1 = (\omega t + \phi_2) - (\omega t + \phi_1 - \frac{\pi}{2})$
$\Delta \Phi = \phi_2 - \phi_1 + \frac{\pi}{2}$

Alternatively, if we convert $x_2$ to a sine function:
$x_2 = A \cos(\omega t + \phi_2) = A \sin((\omega t + \phi_2) + \frac{\pi}{2})$
$x_2 = A \sin(\omega t + \phi_2 + \frac{\pi}{2})$

Now we have:
Phase of $x_1$: $\Phi_1 = \omega t + \phi_1$
Phase of $x_2$: $\Phi_2 = \omega t + \phi_2 + \frac{\pi}{2}$

The phase difference $\Delta \Phi$ is:
$\Delta \Phi = \Phi_2 - \Phi_1 = (\omega t + \phi_2 + \frac{\pi}{2}) - (\omega t + \phi_1)$
$\Delta \Phi = \phi_2 - \phi_1 + \frac{\pi}{2}$

Both methods yield the same result. The phase difference is $(\phi_2 - \phi_1 + \frac{\pi}{2})$.

Question 12

Q12. A particle is executing SHM with amplitude $A$ and angular frequency $\omega$. Find the time taken by the particle to go from $x = 0$ to $x = A/2$.

Accepted Answer

Let the displacement of the particle executing SHM be given by $x(t) = A \sin(\omega t)$. This choice of equation implies that at $t=0$, the particle is at $x=0$ and moving towards positive $x$ (if $\phi=0$).

We need to find the time $t$ when the displacement $x = A/2$.

Substitute $x = A/2$ into the displacement equation:
$A/2 = A \sin(\omega t)$

Divide both sides by $A$:
$1/2 = \sin(\omega t)$

To find $\omega t$, take the inverse sine of $1/2$. The principal value for which $\sin(	heta) = 1/2$ is $	heta = \pi/6$ radians.
$\omega t = \pi/6$

Now, solve for $t$:
$t = \frac{\pi}{6\omega}$

Alternatively, if we use $x(t) = A \cos(\omega t + \phi)$:
If the particle starts from $x=0$ at $t=0$, then $0 = A \cos(\phi)$, which implies $\phi = \pi/2$ (or $3\pi/2$). Let's use $\phi = \pi/2$.
So, $x(t) = A \cos(\omega t + \pi/2) = -A \sin(\omega t)$. This implies motion towards negative $x$ initially. To ensure motion towards positive $x$, we should use $x(t) = A \sin(\omega t)$.

If we use $x(t) = A \cos(\omega t)$ and assume the particle starts from $x=A$ at $t=0$, then we would need to find the time to reach $x=0$ and then $x=A/2$ from $x=0$. This is more complex. The simplest approach is to assume $x=0$ at $t=0$ for the starting point.

So, the time taken for the particle to go from $x = 0$ to $x = A/2$ is $\frac{\pi}{6\omega}$.

Question 13

Q13. A body describes simple harmonic motion with amplitude 5 cm and period 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.

Accepted Answer

Given:
Amplitude, A = 5 cm = 0.05 m
Period, T = 0.2 s

First, calculate the angular frequency (ω):
ω = 2π/T = 2π/0.2 = 10π rad/s

Now, we calculate velocity (v) and acceleration (a) for the given displacements:
Velocity, v = ω√(A² - x²)
Acceleration, a = -ω²x

(a) When displacement x = 5 cm = 0.05 m (at the extreme position):
Velocity, v = 10π√((0.05)² - (0.05)²) = 10π√(0) = 0 m/s
Acceleration, a = -(10π)²(0.05) = -100π²(0.05) = -5π² m/s²

(b) When displacement x = 3 cm = 0.03 m:
Velocity, v = 10π√((0.05)² - (0.03)²) = 10π√(0.0025 - 0.0009) = 10π√(0.0016) = 10π(0.04) = 0.4π m/s
Acceleration, a = -(10π)²(0.03) = -100π²(0.03) = -3π² m/s²

(c) When displacement x = 0 cm = 0 m (at the mean position):
Velocity, v = 10π√((0.05)² - (0)²) = 10π(0.05) = 0.5π m/s
Acceleration, a = -(10π)²(0) = 0 m/s²

Summary of results:
(a) At x = 5 cm: v = 0 m/s, a = -5π² m/s²
(b) At x = 3 cm: v = 0.4π m/s, a = -3π² m/s²
(c) At x = 0 cm: v = 0.5π m/s, a = 0 m/s²

Question 14

Q14. At what distance from the mean position is the speed of a particle performing SHM half its maximum speed? Given path length of SHM is 10 cm.

Accepted Answer

Given:
Path length of SHM = 10 cm

First, determine the amplitude (A):
Amplitude A = Path length / 2 = 10 cm / 2 = 5 cm = 0.05 m

Recall the formulas for speed in SHM:
Speed at displacement x: v = ω√(A² - x²)
Maximum speed (at mean position, x=0): v_max = ωA

We are given that the speed v is half its maximum speed:
v = v_max / 2

Substitute the formulas for v and v_max:
ω√(A² - x²) = (ωA) / 2

Cancel ω from both sides (since ω ≠ 0):
√(A² - x²) = A / 2

Square both sides to eliminate the square root:
A² - x² = (A / 2)²
A² - x² = A² / 4

Rearrange the equation to solve for x²:
x² = A² - A² / 4
x² = (4A² - A²) / 4
x² = 3A² / 4

Take the square root of both sides to find x:
x = ±√(3A² / 4)
x = ±(√3 / 2)A

Now, substitute the value of A = 5 cm:
x = ±(√3 / 2) * 5 cm
x = ±(1.732 / 2) * 5 cm
x = ±0.866 * 5 cm
x = ±4.33 cm

Thus, the speed of the particle is half its maximum speed at a distance of approximately 4.33 cm from the mean position.

Question 15

Q15. What is the period of oscillation of a simple pendulum at a carth's satellite? (or at the centre of the earth?)

Accepted Answer

The period of oscillation of a simple pendulum is given by the formula:
T = 2π√(L/g)
where L is the length of the pendulum and g is the effective acceleration due to gravity.

**1. At a Earth's satellite (in orbit):**
An Earth's satellite is in a state of continuous free fall around the Earth. In this condition, objects inside the satellite experience apparent weightlessness. This means the effective acceleration due to gravity (g_eff) inside the satellite is zero.

If g_eff = 0, then substituting this into the period formula:
T = 2π√(L/0)
This implies T approaches infinity.

Therefore, a simple pendulum will not oscillate in an Earth's satellite. It will simply float in any position, as there is no restoring force due to gravity to bring it back to the mean position.

**2. At the centre of the Earth:**
According to Newton's law of gravitation, the gravitational force inside a uniform spherical shell is zero. For a point mass at the center of a uniform sphere, the net gravitational force due to all parts of the sphere cancels out. Thus, the effective acceleration due to gravity (g_eff) at the center of the Earth is zero.

Similar to the satellite case, if g_eff = 0, then:
T = 2π√(L/0)
This implies T approaches infinity.

Therefore, a simple pendulum will also not oscillate at the center of the Earth. It would remain in any position it is placed, as there is no net gravitational force to act as a restoring force.

Question 16

Q16. A body of mass 1 kg is attached to a spring of force constant 600 N/m, and undergoes SHM. Its maximum velocity is 1.2 m/s. Calculate:
(a) The amplitude of oscillation.
(b) The to

Accepted Answer

Given:
Mass of the body, m = 1 kg
Force constant of the spring, k = 600 N/m
Maximum velocity, v_max = 1.2 m/s

**Step 1: Calculate the angular frequency (ω)**
For a spring-mass system, the angular frequency is given by:
ω = √(k/m)
ω = √(600 N/m / 1 kg)
ω = √600 rad/s
ω ≈ 24.49 rad/s

**(a) The amplitude of oscillation (A)**
We know that the maximum velocity in SHM is given by:
v_max = ωA
Rearranging for A:
A = v_max / ω
A = 1.2 m/s / √600 rad/s
A = 1.2 / 24.49
A ≈ 0.049 m
So, the amplitude of oscillation is approximately 0.049 m or 4.9 cm.

**(b) The total energy of oscillation (E)**
The total energy of oscillation in SHM can be calculated using the maximum kinetic energy or the maximum potential energy. Using maximum kinetic energy:
E = (1/2)mv_max²
E = (1/2) * 1 kg * (1.2 m/s)²
E = (1/2) * 1 * 1.44
E = 0.72 J

Alternatively, using the amplitude and force constant:
E = (1/2)kA²
E = (1/2) * 600 N/m * (0.049 m)²
E = 300 * 0.002401
E ≈ 0.72 J
So, the total energy of oscillation is 0.72 J.

**(c) The period of oscillation (T)**
The period of oscillation is related to the angular frequency by:
T = 2π/ω
T = 2π / √600
T = 2π / 24.49
T ≈ 0.256 s
So, the period of oscillation is approximately 0.256 s.

**Final Answers:**
(a) Amplitude of oscillation, A ≈ 0.049 m (or 4.9 cm)
(b) Total energy of oscillation, E = 0.72 J
(c) Period of oscillation, T ≈ 0.256 s

Question 17

Q17. A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm,

Accepted Answer

Given:
Amplitude, A = 5 cm = 0.05 m
Period, T = 0.2 s

First, calculate the angular frequency (ω):
ω = 2π/T = 2π/0.2 = 10π rad/s

(a) When displacement x = 5 cm = 0.05 m:
This is the extreme position (x = A).
Velocity, v = ω√(A² - x²) = 10π√( (0.05)² - (0.05)² ) = 0 m/s
Acceleration, a = -ω²x = -(10π)²(0.05) = -100π²(0.05) = -5π² m/s²
Using π² ≈ 9.87, a = -5 × 9.87 = -49.35 m/s²

(b) When displacement x = 3 cm = 0.03 m:
Velocity, v = ω√(A² - x²) = 10π√( (0.05)² - (0.03)² )
v = 10π√(0.0025 - 0.0009) = 10π√(0.0016) = 10π(0.04) = 0.4π m/s
Using π ≈ 3.14, v = 0.4 × 3.14 = 1.256 m/s
Acceleration, a = -ω²x = -(10π)²(0.03) = -100π²(0.03) = -3π² m/s²
Using π² ≈ 9.87, a = -3 × 9.87 = -29.61 m/s²

(c) When displacement x = 0 cm = 0 m:
This is the mean position.
Velocity, v = ω√(A² - x²) = 10π√( (0.05)² - 0² ) = 10π(0.05) = 0.5π m/s
Using π ≈ 3.14, v = 0.5 × 3.14 = 1.57 m/s
Acceleration, a = -ω²x = -(10π)²(0) = 0 m/s²

Summary:
(a) At x = 5 cm: Velocity = 0 m/s, Acceleration = -5π² m/s² (or -49.35 m/s²)
(b) At x = 3 cm: Velocity = 0.4π m/s (or 1.256 m/s), Acceleration = -3π² m/s² (or -29.61 m/s²)
(c) At x = 0 cm: Velocity = 0.5π m/s (or 1.57 m/s), Acceleration = 0 m/s²

Question 18

Q18. At what distance from the mean position is the kinetic energy of a particle performing SHM equal to its potential energy?

Accepted Answer

Let the displacement of the particle from the mean position be x.

For a particle performing SHM, the kinetic energy (KE) and potential energy (PE) are given by:
Kinetic Energy, KE = ½ mω²(A² - x²)
Potential Energy, PE = ½ mω²x²
where m is the mass, ω is the angular frequency, A is the amplitude, and x is the displacement.

We are given that KE = PE.
So, ½ mω²(A² - x²) = ½ mω²x²

Divide both sides by ½ mω² (since m and ω are non-zero):
A² - x² = x²
A² = 2x²
x² = A²/2
x = ±√(A²/2)
x = ± A/√2

Thus, the kinetic energy of a particle performing SHM is equal to its potential energy when its displacement from the mean position is A/√2 (or A/√(2)).

Question 19

Q19. A particle executing SHM has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s². Find the period and amplitude of the motion.

Accepted Answer

Given:
Maximum speed, v_max = 30 cm/s = 0.30 m/s
Maximum acceleration, a_max = 60 cm/s² = 0.60 m/s²

For a particle executing SHM, the maximum speed and maximum acceleration are given by:
v_max = Aω  --- (1)
a_max = Aω² --- (2)
where A is the amplitude and ω is the angular frequency.

To find ω, divide equation (2) by equation (1):
a_max / v_max = (Aω²) / (Aω)
a_max / v_max = ω
ω = 60 cm/s² / 30 cm/s = 2 rad/s

Now, calculate the period (T) using ω:
T = 2π/ω = 2π/2 = π s
Using π ≈ 3.14, T = 3.14 s

To find the amplitude (A), substitute ω into equation (1):
v_max = Aω
30 cm/s = A × 2 rad/s
A = 30 cm/s / 2 rad/s = 15 cm

Alternatively, using SI units:
A = 0.30 m/s / 2 rad/s = 0.15 m

Therefore, the period of the motion is π s (or 3.14 s) and the amplitude is 15 cm (or 0.15 m).

Question 20

Q20. As a car passes a speed breaker, it is observed to oscillate up and down with a period of 0.5 s. The amplitude of oscillation is 10 cm. If the total mass of the car and its occupan

Accepted Answer

Given:
Period of oscillation, T = 0.5 s
Amplitude of oscillation, A = 10 cm = 0.10 m (Note: Amplitude is not needed for calculating spring constant from period and mass)
Total mass of the car and occupants, m = 1000 kg

For a mass-spring system, the period of oscillation is given by:
T = 2π√(m/k)
where m is the mass and k is the spring constant.

We need to find k. Rearrange the formula:
T² = (2π)² (m/k)
T² = 4π² (m/k)
k = 4π²m / T²

Substitute the given values:
k = (4 × (3.14159)² × 1000 kg) / (0.5 s)²
k = (4 × 9.8696 × 1000) / 0.25
k = (39.4784 × 1000) / 0.25
k = 39478.4 / 0.25
k = 157913.6 N/m

Rounding to appropriate significant figures, k ≈ 1.58 × 10⁵ N/m.

Therefore, the spring constant of the suspension system is approximately 1.58 × 10⁵ N/m.

Question 21

Q21. A simple pendulum of length 1 m has a bob of mass 100 g. It is displaced by 2 cm from its mean position and released. Calculate (a) the period of oscillation, (b) the maximum veloc

Accepted Answer

Given:
Length of the simple pendulum, L = 1 m
Mass of the bob, m = 100 g = 0.1 kg
Amplitude (maximum displacement), A = 2 cm = 0.02 m
Acceleration due to gravity, g = 9.8 m/s² (standard value)

(a) The period of oscillation (T):
For a simple pendulum, the period is given by T = 2π√(L/g).
T = 2π√(1 m / 9.8 m/s²)
T = 2π√(0.10204)
T = 2π × 0.3194
T ≈ 2.007 s

(b) The maximum velocity of the bob (v_max):
First, calculate the angular frequency (ω):
ω = 2π/T = 2π / 2.007 s ≈ 3.13 rad/s
Alternatively, ω = √(g/L) = √(9.8/1) = √9.8 ≈ 3.13 rad/s
Maximum velocity, v_max = Aω
v_max = 0.02 m × 3.13 rad/s
v_max ≈ 0.0626 m/s

(c) The maximum acceleration of the bob (a_max):
Maximum acceleration, a_max = Aω²
a_max = 0.02 m × (3.13 rad/s)²
a_max = 0.02 × 9.7969
a_max ≈ 0.1959 m/s²

(d) The total energy of the simple pendulum (E):
For SHM, the total energy is given by E = ½ mω²A².
E = ½ × 0.1 kg × (3.13 rad/s)² × (0.02 m)²
E = ½ × 0.1 × 9.7969 × 0.0004
E = 0.05 × 9.7969 × 0.0004
E ≈ 0.0001959 J
E ≈ 1.96 × 10⁻⁴ J

Summary:
(a) Period of oscillation, T ≈ 2.01 s
(b) Maximum velocity of the bob, v_max ≈ 0.063 m/s
(c) Maximum acceleration of the bob, a_max ≈ 0.196 m/s²
(d) Total energy of the simple pendulum, E ≈ 1.96 × 10⁻⁴ J

Question 22

Q22. A body of mass 10 kg is attached to a spring of spring constant 600 N m–1. It is stretched by a distance of 5 cm from its equilibrium position and released. Find (a) the angular fr

Accepted Answer

Given:
Mass of the body, m = 10 kg
Spring constant, k = 600 N m⁻¹
Initial stretch (amplitude), A = 5 cm = 0.05 m

(a) Angular frequency (ω):
ω = √(k/m) = √(600 N m⁻¹ / 10 kg) = √(60) rad/s ≈ 7.746 rad/s

(b) Period (T):
T = 2π/ω = 2π / 7.746 s ≈ 0.811 s

(c) Frequency (f):
f = 1/T = 1 / 0.811 Hz ≈ 1.233 Hz

(d) Amplitude (A):
The amplitude is the maximum displacement from the equilibrium position, which is the initial stretch.
A = 5 cm = 0.05 m

(e) Maximum speed (v_max):
v_max = Aω = 0.05 m × 7.746 rad/s ≈ 0.3873 m/s

(f) Position for maximum speed:
The speed of the body is maximum at the equilibrium position (x = 0).

(g) Maximum acceleration (a_max):
a_max = Aω² = 0.05 m × (7.746 rad/s)² = 0.05 m × 60 rad²/s² = 3 m/s²

(h) Position for maximum acceleration:
The acceleration of the body is maximum at the extreme positions (x = +A and x = -A).

Question 23

Q23. A body of mass 5 kg is attached to a spring of spring constant 500 N m–1. It is stretched by a distance of 10 cm from its equilibrium position and released. Find (a) the angular fr

Accepted Answer

Given:
Mass of the body, m = 5 kg
Spring constant, k = 500 N m⁻¹
Initial stretch (amplitude), A = 10 cm = 0.10 m

(a) Angular frequency (ω):
ω = √(k/m) = √(500 N m⁻¹ / 5 kg) = √(100) rad/s = 10 rad/s

(b) Period (T):
T = 2π/ω = 2π / 10 s = π/5 s ≈ 0.628 s

(c) Frequency (f):
f = 1/T = 1 / (π/5) Hz = 5/π Hz ≈ 1.592 Hz

(d) Amplitude (A):
The amplitude is the maximum displacement from the equilibrium position, which is the initial stretch.
A = 10 cm = 0.10 m

(e) Maximum speed (v_max):
v_max = Aω = 0.10 m × 10 rad/s = 1 m/s

(f) Position for maximum speed:
The speed of the body is maximum at the equilibrium position (x = 0).

(g) Maximum acceleration (a_max):
a_max = Aω² = 0.10 m × (10 rad/s)² = 0.10 m × 100 rad²/s² = 10 m/s²

(h) Position for maximum acceleration:
The acceleration of the body is maximum at the extreme positions (x = +A and x = -A).

Question 24

Q24. A body of mass 2 kg is attached to a spring of spring constant 200 N m–1. It is stretched by a distance of 20 cm from its equilibrium position and released. Find (a) the angular fr

Accepted Answer

Given:
Mass of the body, m = 2 kg
Spring constant, k = 200 N m⁻¹
Initial stretch (amplitude), A = 20 cm = 0.20 m

(a) Angular frequency (ω):
ω = √(k/m) = √(200 N m⁻¹ / 2 kg) = √(100) rad/s = 10 rad/s

(b) Period (T):
T = 2π/ω = 2π / 10 s = π/5 s ≈ 0.628 s

(c) Frequency (f):
f = 1/T = 1 / (π/5) Hz = 5/π Hz ≈ 1.592 Hz

(d) Amplitude (A):
The amplitude is the maximum displacement from the equilibrium position, which is the initial stretch.
A = 20 cm = 0.20 m

(e) Maximum speed (v_max):
v_max = Aω = 0.20 m × 10 rad/s = 2 m/s

(f) Position for maximum speed:
The speed of the body is maximum at the equilibrium position (x = 0).

(g) Maximum acceleration (a_max):
a_max = Aω² = 0.20 m × (10 rad/s)² = 0.20 m × 100 rad²/s² = 20 m/s²

(h) Position for maximum acceleration:
The acceleration of the body is maximum at the extreme positions (x = +A and x = -A).

Question 25

Q25. A body of mass 0.5 kg is attached to a spring of spring constant 100 N m–1. It is stretched by a distance of 15 cm from its equilibrium position and released. Find (a) the angular

Accepted Answer

Given:
Mass of the body, m = 0.5 kg
Spring constant, k = 100 N m⁻¹
Initial stretch (amplitude), A = 15 cm = 0.15 m

(a) Angular frequency (ω):
ω = √(k/m) = √(100 N m⁻¹ / 0.5 kg) = √(200) rad/s ≈ 14.142 rad/s

(b) Period (T):
T = 2π/ω = 2π / 14.142 s ≈ 0.444 s

(c) Frequency (f):
f = 1/T = 1 / 0.444 Hz ≈ 2.252 Hz

(d) Amplitude (A):
The amplitude is the maximum displacement from the equilibrium position, which is the initial stretch.
A = 15 cm = 0.15 m

(e) Maximum speed (v_max):
v_max = Aω = 0.15 m × 14.142 rad/s ≈ 2.1213 m/s

(f) Position for maximum speed:
The speed of the body is maximum at the equilibrium position (x = 0).

(g) Maximum acceleration (a_max):
a_max = Aω² = 0.15 m × (14.142 rad/s)² = 0.15 m × 200 rad²/s² = 30 m/s²

(h) Position for maximum acceleration:
The acceleration of the body is maximum at the extreme positions (x = +A and x = -A).

Chapter 14: Oscillations