Q: Q12. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in a direction from the yard to the station with a speed of 10 m s⁻¹.

For the listener on the platform:\n- The frequency of sound heard is 400 Hz.\n- The speed of sound (effective) is 350 m/s.\n- The wavelength of the sound is 0.875 m.\n\nThe effect of the wind is NOT equivalent to the train moving towards or away from the listener. Wind affects the speed of sound relative to the ground, but not the relative motion between the source and the medium at the source, or the observer and the medium at the observer.

Question 1

Q1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the dis

Accepted Answer

The time taken for the disturbance to reach the other end of the string is 0.50 seconds.

Question 2

Q2. A stone dropped from the top of a tower of height 200 m splashes into the water of a pond. Assuming the speed of sound in air to be 340 m s⁻¹, when will the splash be heard at the

Accepted Answer

The splash will be heard at the top of the tower approximately 6.79 seconds after the stone is dropped.

Question 3

Q3. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dr

Accepted Answer

The tension in the wire should be approximately 2076.68 N.

Question 4

Q4. Use the formula v = √(γP/ρ) to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.

Accepted Answer

The speed of sound in air, given by the formula v = √(γP/ρ), is influenced by various factors. Let's analyze how pressure, temperature, and humidity affect it.

Question 5

Q5. You may have observed that a cricket commentator often uses the expression "the pitch was lively today". What does the expression mean?

Accepted Answer

In the context of cricket, when a commentator says "the pitch was lively today", it refers to the condition of the cricket pitch (the central strip of the ground where the bowling and batting take place). A 'lively' pitch is one that is hard, dry, and has good bounce. This type of pitch causes the cricket ball to bounce higher and faster after hitting the surface, making it more challenging for the batsmen to play shots. It generally favors fast bowlers who can exploit the extra bounce and pace to trouble the batsmen.

Question 6

Q6. A string fixed at both ends is vibrating in the lowest mode of vibration (in 1st harmonic) with a frequency of 200 Hz. The maximum displacement of the mid-point of the string is 0.

Accepted Answer

Here's a detailed solution for the given problem:

**Given:**
*   Frequency (f) = 200 Hz
*   Maximum displacement at mid-point (A_max_mid) = 0.5 cm = 0.005 m
*   Length of the string (L) = 1.5 m
*   Mode of vibration = Lowest mode (1st harmonic)

**(a) Wavelength and Speed of the Transverse Wave**
For the lowest mode (1st harmonic) of a string fixed at both ends, the string vibrates with one loop. In this mode, the length of the string is equal to half the wavelength (L = λ/2).

Wavelength (λ) = 2L = 2 × 1.5 m = 3.0 m

The speed of the transverse wave (v) is given by v = fλ.
v = 200 Hz × 3.0 m = 600 m/s

**(b) Amplitude of the particle at x = 0.25 m from one end**
The equation of a stationary wave on a string fixed at both ends can be written as:
y(x, t) = (2A) sin(kx) cos(ωt)
where 2A is the maximum displacement (amplitude) at an antinode.

At the mid-point (x = L/2 = 1.5/2 = 0.75 m), the displacement is maximum, which means it's an antinode. The maximum displacement at the mid-point is given as 0.5 cm.
So, 2A = 0.5 cm = 0.005 m.

The wave number (k) = 2π/λ = 2π / 3.0 m = (2π/3) rad/m.

The amplitude of a particle at position x is given by A_particle(x) = (2A) sin(kx).
We need to find the amplitude at x = 0.25 m.
A_particle(0.25 m) = (0.005 m) sin((2π/3) × 0.25)
A_particle(0.25 m) = (0.005 m) sin(π/6)
A_particle(0.25 m) = (0.005 m) × (1/2)
A_particle(0.25 m) = 0.0025 m = 0.25 cm

**(c) Maximum velocity and acceleration of the particle at the mid-point of the string**
The angular frequency (ω) = 2πf = 2π × 200 Hz = 400π rad/s.

At the mid-point, the particle undergoes Simple Harmonic Motion (SHM) with an amplitude equal to the maximum displacement at the mid-point, A_mid = 0.5 cm = 0.005 m.

Maximum velocity (v_max) of a particle in SHM = A_mid × ω
v_max = 0.005 m × 400π rad/s
v_max = 2π m/s ≈ 6.28 m/s

Maximum acceleration (a_max) of a particle in SHM = A_mid × ω²
a_max = 0.005 m × (400π rad/s)²
a_max = 0.005 m × 160000π² rad²/s²
a_max = 800π² m/s²
a_max ≈ 800 × (3.14159)² m/s² ≈ 800 × 9.8696 m/s² ≈ 7895.68 m/s²

**(d) Equation of the stationary wave**
The general equation for a stationary wave is y(x, t) = (2A) sin(kx) cos(ωt).

From previous calculations:
*   2A = 0.005 m
*   k = 2π/3 rad/m
*   ω = 400π rad/s

Substituting these values:
y(x, t) = 0.005 sin((2π/3)x) cos(400πt)

Therefore, the equation of the stationary wave is y(x, t) = 0.005 sin((2π/3)x) cos(400πt), where y and x are in meters and t is in seconds.

Question 7

Q7. A string fixed at both ends has a length of 1.0 m and a mass of 5.0 × 10⁻² kg. The string is under a tension of 80 N.
(a) Calculate the speed of transverse waves on the string.
(

Accepted Answer

Here's a detailed solution for the given problem:

**Given:**
*   Length of the string (L) = 1.0 m
*   Mass of the string (M) = 5.0 × 10⁻² kg = 0.05 kg
*   Tension in the string (T) = 80 N

**(a) Speed of transverse waves on the string**
First, calculate the linear mass density (μ) of the string:
μ = M / L = 0.05 kg / 1.0 m = 0.05 kg/m

The speed of transverse waves (v) on a string is given by the formula:
v = √(T/μ)
v = √(80 N / 0.05 kg/m)
v = √(1600 m²/s²)
v = 40 m/s

**(b) Frequency of the fundamental mode of vibration**
For a string fixed at both ends, the fundamental frequency (f₁) (or 1st harmonic) is given by:
f₁ = v / (2L)
f₁ = 40 m/s / (2 × 1.0 m)
f₁ = 40 / 2 Hz
f₁ = 20 Hz

**(c) Frequencies of the first two overtones**
For a string fixed at both ends, all integer multiples of the fundamental frequency are possible harmonics.
*   The first overtone is the 2nd harmonic (f₂ = 2f₁).
*   The second overtone is the 3rd harmonic (f₃ = 3f₁).

First overtone (f₂) = 2 × f₁ = 2 × 20 Hz = 40 Hz
Second overtone (f₃) = 3 × f₁ = 3 × 20 Hz = 60 Hz

**(d) Harmonics absent if the string is plucked at a distance of 0.25 m from one end**
When a string is plucked at a certain point, the harmonics that have a node at that point will be absent in the resulting vibration. The plucked point acts as an antinode for the excited harmonics, but it cannot be an antinode for harmonics that require a node at that position.

The positions of nodes for the nth harmonic are given by x = (m/n)L, where m = 0, 1, 2, ..., n.
Alternatively, a simpler rule is: if the string is plucked at a distance L/p from one end, then the p-th harmonic and its multiples (2p, 3p, etc.) will be absent.

Given plucking distance = 0.25 m.
L = 1.0 m.
So, the plucking distance is L/4 (since 0.25 m = 1.0 m / 4).

Therefore, the 4th harmonic and its integer multiples (8th, 12th, 16th, etc.) will be absent.

Absent harmonics: 4th, 8th, 12th, 16th, ... (i.e., all harmonics that are multiples of 4).

Question 8

Q8. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends ar

Accepted Answer

Here's a detailed solution for the given problem:

**Given:**
*   Length of the pipe (L) = 20 cm = 0.20 m
*   Frequency of the source (f_source) = 430 Hz
*   Speed of sound in air (v) = 340 m/s

**Part 1: Pipe closed at one end**
For a pipe closed at one end, only odd harmonics are possible. The frequencies of the harmonics are given by:
f_n = n * (v / 4L), where n = 1, 3, 5, ...

First, calculate the fundamental frequency (n=1):
f₁ = 1 * (340 m/s / (4 × 0.20 m))
f₁ = 340 m/s / 0.80 m
f₁ = 425 Hz

Now, let's find which harmonic mode (n) corresponds to the source frequency of 430 Hz. We can use the formula f_n = n * f₁:
430 Hz = n * 425 Hz
n = 430 / 425 ≈ 1.01

Since n must be an odd integer for a closed pipe, and 1.01 is very close to 1, it implies that the 1st harmonic (fundamental mode) is resonantly excited. The slight difference (430 Hz vs 425 Hz) suggests that the source is very close to exciting the fundamental mode, or there might be a slight discrepancy in the given values or ideal conditions. However, among the available odd harmonics, the 1st harmonic (425 Hz) is the closest to 430 Hz. If we consider practical resonance, the 1st harmonic would be excited.

*Self-correction/Clarification for NEET:* In NEET, if the calculated frequency is very close to a harmonic, that harmonic is considered to be excited. Here, 430 Hz is very close to 425 Hz (1st harmonic). If the question implies exact resonance, then strictly speaking, 430 Hz does not exactly match any harmonic of this pipe. However, if we are looking for the 'closest' harmonic, it's the 1st harmonic.

Let's assume the question implies the closest harmonic or that 430 Hz is intended to excite the fundamental.
**Conclusion for closed pipe:** The 1st harmonic (fundamental mode) is resonantly excited by the 430 Hz source (assuming practical resonance or slight tolerance in values).

**Part 2: Pipe open at both ends**
For a pipe open at both ends, all integer harmonics are possible. The frequencies of the harmonics are given by:
f_n' = n * (v / 2L), where n = 1, 2, 3, ...

First, calculate the fundamental frequency (n=1) for the open pipe:
f₁' = 1 * (340 m/s / (2 × 0.20 m))
f₁' = 340 m/s / 0.40 m
f₁' = 850 Hz

Now, let's check if the 430 Hz source will be in resonance with this open pipe. The possible frequencies for the open pipe are multiples of 850 Hz:
f₁' = 850 Hz
f₂' = 2 × 850 Hz = 1700 Hz
...

Since 430 Hz is not equal to 850 Hz or any of its integer multiples, the 430 Hz source will **not** be in resonance with the pipe if both ends are open.

**Final Answer Summary:**
*   For the pipe closed at one end, the 1st harmonic (fundamental mode) is resonantly excited by the 430 Hz source (as 425 Hz is very close to 430 Hz).
*   If the pipe is open at both ends, the 430 Hz source will not be in resonance with the pipe.

Question 9

Q9. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the dis

Accepted Answer

The time taken for the disturbance to reach the other end of the string is 0.50 seconds.

Question 10

Q10. A stone dropped from the top of a tower of height 200 m splashes into the water of a pond at the base of the tower. When is the splash heard at the top? Given, speed of sound = 340

Accepted Answer

The splash is heard at the top of the tower approximately 6.79 seconds after the stone is dropped.

Question 11

Q11. A train stands at a platform blowing a whistle of frequency 400 Hz in still air.
(a) What is the frequency of the whistle heard by a man standing on the platform near the train?

Accepted Answer

(a) The frequency heard by the man on the platform is 400 Hz.
(b) The frequency heard by the man inside the train is 400 Hz.
(c) The frequency heard by the man standing on the platform is approximately 388.6 Hz.
(d) The speed of sound in still air is typically taken as 340 m/s (as used in calculations).
(e) The frequency heard by the man inside the train is 400 Hz.

Question 12

Q12. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in a direction from the yard to the station with a speed of 10 m s⁻¹.

Accepted Answer

For the listener on the platform:
- The frequency of sound heard is 400 Hz.
- The speed of sound (effective) is 350 m/s.
- The wavelength of the sound is 0.875 m.

The effect of the wind is NOT equivalent to the train moving towards or away from the listener. Wind affects the speed of sound relative to the ground, but not the relative motion between the source and the medium at the source, or the observer and the medium at the observer.

Question 13

Q13. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends ar

Accepted Answer

For a pipe closed at one end, the fundamental frequency is 425 Hz. The given source frequency of 430 Hz is very close to the fundamental frequency, so it excites the first harmonic (fundamental mode). If the pipe is open at both ends, the fundamental frequency would be 850 Hz. Since the source frequency (430 Hz) is not a harmonic of 850 Hz, it will not be in resonance with the pipe if both ends are open.

Question 14

Q14. A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
(a) What is the frequency of the whistle for a platform observer when

Accepted Answer

(a) (i) When the train approaches the platform, the observed frequency is 412.12 Hz.
(ii) When the train recedes from the platform, the observed frequency is 388.57 Hz.
(b) The speed of sound in still air remains 340 m/s in both cases, as the speed of sound is independent of the motion of the source or observer relative to the medium.

Question 15

Q15. A travelling harmonic wave on a string is described by
y(x, t) = 7.5 sin (0.0050 x + 12 t + π/4)
(a) What are the propagation direction, wavelength, frequency, and amplitude of t

Accepted Answer

(a) The wave propagates in the negative x-direction. The amplitude is 7.5 cm, wavelength is 1256.6 cm (or 12.566 m), and frequency is 1.91 Hz.
(b) At x = 1 cm and t = 1 s, the transverse speed of a point on the string is -74.9 cm/s and the transverse acceleration is -898.8 cm/s².
(c) The speed of wave propagation is 2400 cm/s (or 24 m/s).

Question 16

Q16. A string fixed at both ends is vibrating in the n-th overtone. The distance between two consecutive nodes is 2.0 cm. The length of the string is 10.0 cm. Find the order of the over

Accepted Answer

The order of the overtone is 4.

Question 17

Q17. A string fixed at both ends is vibrating in the n-th mode of vibration. Is the amplitude at an antinode equal to the amplitude at an adjacent node? Explain your answer.

Accepted Answer

No, the amplitude at an antinode is NOT equal to the amplitude at an adjacent node.

**Explanation:**
When a string fixed at both ends vibrates, it forms standing waves. A standing wave is formed by the superposition of two identical waves traveling in opposite directions.

1.  **Nodes:** These are points on the string where the displacement (and hence amplitude) is always zero. At nodes, the two component waves interfere destructively at all times.
2.  **Antinodes:** These are points on the string where the displacement amplitude is maximum. At antinodes, the two component waves interfere constructively, resulting in the largest possible displacement from the equilibrium position.

Therefore, by definition, the amplitude at a node is zero, while the amplitude at an antinode is maximum. They are fundamentally different and not equal.

Question 18

Q18. A string fixed at both ends vibrates in its fundamental mode with a frequency of 440 Hz. The length of the string is 0.5 m.
(a) Determine the speed of transverse waves on the stri

Accepted Answer

Here's the solution for the given problem:

**(a) Speed of transverse waves on the string:**
For a string fixed at both ends, the frequency of the fundamental mode (first harmonic, n=1) is given by:
$f_1 = \frac{v}{2L}$
Where:
$f_1$ = fundamental frequency = 440 Hz
$L$ = length of the string = 0.5 m
$v$ = speed of transverse waves on the string

Rearranging the formula to find $v$:
$v = 2L f_1$
$v = 2 	imes 0.5 	ext{ m} 	imes 440 	ext{ Hz}$
$v = 1 	ext{ m} 	imes 440 	ext{ Hz}$
$v = 440 	ext{ m/s}$

**The speed of transverse waves on the string is 440 m/s.**

**(b) New frequency of vibration in the second harmonic:**
The frequency of the n-th harmonic for a string fixed at both ends is given by:
$f_n = n f_1$
For the second harmonic, n = 2.
$f_2 = 2 f_1$
$f_2 = 2 	imes 440 	ext{ Hz}$
$f_2 = 880 	ext{ Hz}$

**The new frequency of vibration in the second harmonic is 880 Hz.**

**(c) Wavelength of the waves in the string when it vibrates in its second harmonic:**
For the n-th harmonic, the wavelength $\lambda_n$ is given by:
$\lambda_n = \frac{2L}{n}$
For the second harmonic, n = 2.
$\lambda_2 = \frac{2L}{2} = L$
$\lambda_2 = 0.5 	ext{ m}$

Alternatively, using the wave equation $v = f \lambda$:
$\lambda_2 = \frac{v}{f_2}$
$\lambda_2 = \frac{440 	ext{ m/s}}{880 	ext{ Hz}}$
$\lambda_2 = 0.5 	ext{ m}$

**The wavelength of the waves in the string when it vibrates in its second harmonic is 0.5 m.**

Question 19

Q19. A string of length 1.5 m with its ends fixed is vibrating in its fundamental mode. The speed of transverse waves on the string is 60 m/s.
(a) What is the frequency of vibration?

Accepted Answer

Here's the solution for the given problem:

**Given:**
Length of the string, $L = 1.5 	ext{ m}$
Speed of transverse waves, $v = 60 	ext{ m/s}$

**(a) Frequency of vibration (fundamental mode):**
For a string fixed at both ends vibrating in its fundamental mode (first harmonic, n=1), the frequency ($f_1$) is given by:
$f_1 = \frac{v}{2L}$
Substitute the given values:
$f_1 = \frac{60 	ext{ m/s}}{2 	imes 1.5 	ext{ m}}$
$f_1 = \frac{60 	ext{ m/s}}{3.0 	ext{ m}}$
$f_1 = 20 	ext{ Hz}$

**The frequency of vibration in the fundamental mode is 20 Hz.**

**(b) Wavelength of the waves in the string (fundamental mode):**
For the fundamental mode (n=1) of a string fixed at both ends, the wavelength ($\lambda_1$) is given by:
$\lambda_1 = \frac{2L}{1} = 2L$
Substitute the length:
$\lambda_1 = 2 	imes 1.5 	ext{ m}$
$\lambda_1 = 3.0 	ext{ m}$

Alternatively, using the wave equation $v = f \lambda$:
$\lambda_1 = \frac{v}{f_1} = \frac{60 	ext{ m/s}}{20 	ext{ Hz}} = 3.0 	ext{ m}$

**The wavelength of the waves in the string in the fundamental mode is 3.0 m.**

**(c) New frequency and wavelength in the third harmonic:**
For the third harmonic, n = 3.

**New frequency ($f_3$):**
The frequency of the n-th harmonic is $f_n = n f_1$.
$f_3 = 3 f_1$
$f_3 = 3 	imes 20 	ext{ Hz}$
$f_3 = 60 	ext{ Hz}$

**The new frequency in the third harmonic is 60 Hz.**

**New wavelength ($\lambda_3$):**
The wavelength of the n-th harmonic is $\lambda_n = \frac{2L}{n}$.
$\lambda_3 = \frac{2L}{3}$
$\lambda_3 = \frac{2 	imes 1.5 	ext{ m}}{3}$
$\lambda_3 = \frac{3.0 	ext{ m}}{3}$
$\lambda_3 = 1.0 	ext{ m}$

Alternatively, using the wave equation $v = f \lambda$:
$\lambda_3 = \frac{v}{f_3} = \frac{60 	ext{ m/s}}{60 	ext{ Hz}} = 1.0 	ext{ m}$

**The new wavelength in the third harmonic is 1.0 m.**

Question 20

Q20. A pipe 30.0 cm long is open at both ends.
(a) Which harmonic mode of the pipe resonates with a 1.1 kHz source?
(b) If one end of the pipe is closed, which harmonic mode of the pi

Accepted Answer

Here's the solution for the given problem:

**Given:**
Length of the pipe, $L = 30.0 	ext{ cm} = 0.30 	ext{ m}$
Source frequency, $f_{source} = 1.1 	ext{ kHz} = 1.1 	imes 10^3 	ext{ Hz} = 1100 	ext{ Hz}$
Speed of sound in air, $v = 330 	ext{ m/s}$

**(a) Harmonic mode for a pipe open at both ends:**
For a pipe open at both ends, the frequencies of the harmonic modes are given by:
$f_n = n \frac{v}{2L}$, where $n = 1, 2, 3, \dots$ (all harmonics are present)

We need to find $n$ such that $f_n = f_{source}$.
$n = \frac{f_{source} 	imes 2L}{v}$

Substitute the given values:
$n = \frac{1100 	ext{ Hz} 	imes 2 	imes 0.30 	ext{ m}}{330 	ext{ m/s}}$
$n = \frac{1100 	imes 0.60}{330}$
$n = \frac{660}{330}$
$n = 2$

Since $n=2$ is an integer, the pipe resonates in its **second harmonic mode** (or first overtone).

**(b) Harmonic mode for a pipe closed at one end:**
If one end of the pipe is closed, the frequencies of the harmonic modes are given by:
$f_n = n \frac{v}{4L}$, where $n = 1, 3, 5, \dots$ (only odd harmonics are present)

We need to find $n$ such that $f_n = f_{source}$.
$n = \frac{f_{source} 	imes 4L}{v}$

Substitute the given values:
$n = \frac{1100 	ext{ Hz} 	imes 4 	imes 0.30 	ext{ m}}{330 	ext{ m/s}}$
$n = \frac{1100 	imes 1.20}{330}$
$n = \frac{1320}{330}$
$n = 4$

Since $n=4$ is an even number, and only odd harmonics are possible for a closed pipe, the pipe **will not resonate** with the 1.1 kHz source when one end is closed. The calculated $n=4$ means that the 1.1 kHz frequency corresponds to the 4th multiple of the fundamental frequency of the closed pipe, but the 4th harmonic is not allowed in a closed pipe (only 1st, 3rd, 5th, etc., are allowed).

Question 21

Q21. A string fixed at both ends is vibrating in the lowest mode of vibration (in 1 loop). The tension in the string is 16 N, mass per unit length is 0.01 kg/m, and the length of the st

Accepted Answer

The problem describes a string fixed at both ends, vibrating in its lowest mode (fundamental mode). We need to calculate the speed of the transverse wave and the frequency of this mode of vibration.

**(a) Speed of the transverse wave on the string:**
The speed of a transverse wave on a string is given by the formula:
v = √(T/μ)
Where:
T = Tension in the string = 16 N
μ = Mass per unit length = 0.01 kg/m

v = √(16 N / 0.01 kg/m) = √(1600 m²/s²) = 40 m/s

**(b) Frequency of the lowest mode of vibration:**
For a string fixed at both ends, the frequency of the nth harmonic (or n loops) is given by:
f_n = n * (v / 2L)
Where:
n = Number of loops (harmonic number)
v = Speed of the transverse wave = 40 m/s
L = Length of the string = 1 m

For the lowest mode of vibration (fundamental mode), n = 1.

f_1 = 1 * (40 m/s / (2 * 1 m)) = 40 m/s / 2 m = 20 Hz

Therefore, the speed of the transverse wave is 40 m/s, and the frequency of the lowest mode of vibration is 20 Hz.

Question 22

Q22. A string fixed at both ends is vibrating in its third overtone (4 loops). The tension in the string is 16 N, mass per unit length is 0.01 kg/m, and the length of the string is 1 m.

Accepted Answer

The problem describes a string fixed at both ends, vibrating in its third overtone. We need to calculate the speed of the transverse wave and the frequency of this mode of vibration.

**(a) Speed of the transverse wave on the string:**
The speed of a transverse wave on a string is given by the formula:
v = √(T/μ)
Where:
T = Tension in the string = 16 N
μ = Mass per unit length = 0.01 kg/m

v = √(16 N / 0.01 kg/m) = √(1600 m²/s²) = 40 m/s

**(b) Frequency of the third overtone:**
For a string fixed at both ends, the frequency of the nth harmonic (or n loops) is given by:
f_n = n * (v / 2L)
Where:
n = Number of loops (harmonic number)
v = Speed of the transverse wave = 40 m/s
L = Length of the string = 1 m

The third overtone corresponds to the 4th harmonic, meaning n = 4 (4 loops).

f_4 = 4 * (40 m/s / (2 * 1 m)) = 4 * (40 m/s / 2 m) = 4 * 20 Hz = 80 Hz

Therefore, the speed of the transverse wave is 40 m/s, and the frequency of the third overtone is 80 Hz.

Question 23

Q23. A string fixed at both ends is vibrating in its second overtone (3 loops). The tension in the string is 16 N, mass per unit length is 0.01 kg/m, and the length of the string is 1 m

Accepted Answer

The problem describes a string fixed at both ends, vibrating in its second overtone. We need to calculate the speed of the transverse wave and the frequency of this mode of vibration.

**(a) Speed of the transverse wave on the string:**
The speed of a transverse wave on a string is given by the formula:
v = √(T/μ)
Where:
T = Tension in the string = 16 N
μ = Mass per unit length = 0.01 kg/m

v = √(16 N / 0.01 kg/m) = √(1600 m²/s²) = 40 m/s

**(b) Frequency of the second overtone:**
For a string fixed at both ends, the frequency of the nth harmonic (or n loops) is given by:
f_n = n * (v / 2L)
Where:
n = Number of loops (harmonic number)
v = Speed of the transverse wave = 40 m/s
L = Length of the string = 1 m

The second overtone corresponds to the 3rd harmonic, meaning n = 3 (3 loops).

f_3 = 3 * (40 m/s / (2 * 1 m)) = 3 * (40 m/s / 2 m) = 3 * 20 Hz = 60 Hz

Therefore, the speed of the transverse wave is 40 m/s, and the frequency of the second overtone is 60 Hz.

Question 24

Q24. A string fixed at both ends is vibrating in its first overtone (2 loops). The tension in the string is 16 N, mass per unit length is 0.01 kg/m, and the length of the string is 1 m.

Accepted Answer

The problem describes a string fixed at both ends, vibrating in its first overtone. We need to calculate the speed of the transverse wave and the frequency of this mode of vibration.

**(a) Speed of the transverse wave on the string:**
The speed of a transverse wave on a string is given by the formula:
v = √(T/μ)
Where:
T = Tension in the string = 16 N
μ = Mass per unit length = 0.01 kg/m

v = √(16 N / 0.01 kg/m) = √(1600 m²/s²) = 40 m/s

**(b) Frequency of the first overtone:**
For a string fixed at both ends, the frequency of the nth harmonic (or n loops) is given by:
f_n = n * (v / 2L)
Where:
n = Number of loops (harmonic number)
v = Speed of the transverse wave = 40 m/s
L = Length of the string = 1 m

The first overtone corresponds to the 2nd harmonic, meaning n = 2 (2 loops).

f_2 = 2 * (40 m/s / (2 * 1 m)) = 2 * (40 m/s / 2 m) = 2 * 20 Hz = 40 Hz

Therefore, the speed of the transverse wave is 40 m/s, and the frequency of the first overtone is 40 Hz.

Question 25

Q25. A string fixed at both ends is vibrating in its fundamental mode (1 loop). The tension in the string is 16 N, mass per unit length is 0.01 kg/m, and the length of the string is 1 m

Accepted Answer

The problem describes a string fixed at both ends, vibrating in its fundamental mode. We need to calculate the speed of the transverse wave and the frequency of this mode of vibration.

**(a) Speed of the transverse wave on the string:**
The speed of a transverse wave on a string is given by the formula:
v = √(T/μ)
Where:
T = Tension in the string = 16 N
μ = Mass per unit length = 0.01 kg/m

v = √(16 N / 0.01 kg/m) = √(1600 m²/s²) = 40 m/s

**(b) Frequency of the fundamental mode:**
For a string fixed at both ends, the frequency of the nth harmonic (or n loops) is given by:
f_n = n * (v / 2L)
Where:
n = Number of loops (harmonic number)
v = Speed of the transverse wave = 40 m/s
L = Length of the string = 1 m

For the fundamental mode, n = 1.

f_1 = 1 * (40 m/s / (2 * 1 m)) = 40 m/s / 2 m = 20 Hz

Therefore, the speed of the transverse wave is 40 m/s, and the frequency of the fundamental mode is 20 Hz.

Question 26

Q26. A string fixed at both ends is vibrating in the third overtone with a frequency of 1200 Hz. The length of the string is 1.5 m.
(a) What is the speed of transverse waves on the str

Accepted Answer

The speed of transverse waves on the string is 300 m/s. The string will vibrate in its fundamental mode at a frequency of 200 Hz.

Question 27

Q27. A closed organ pipe 70 cm long is sounded. If the speed of sound is 330 m/s, what is the fundamental frequency of vibration of the air column? What is the frequency of the first ov

Accepted Answer

The fundamental frequency of the closed organ pipe is 117.86 Hz. The frequency of the first overtone is 353.58 Hz.

Question 28

Q28. A train, standing at the outer signal of a railway station, blows a whistle of frequency 400 Hz in still air.
(a) What is the frequency of the whistle for a platform observer when

Accepted Answer

(a) (i) When the train approaches, the observed frequency is 412.12 Hz. (ii) When the train recedes, the observed frequency is 388.24 Hz.
(b) The speed of sound in each case remains 340 m/s, as it depends only on the medium.
(c) (i) When approaching, the wavelength is 0.825 m. (ii) When receding, the wavelength is 0.875 m.

Question 29

Q29. A train, standing in a station yard, blows a whistle of frequency 800 Hz. The wind is blowing steadily at 10 m/s along the direction of motion of the train towards the station.
(a

Accepted Answer

(a) The frequency of sound heard by the observer on the platform is 800 Hz.
(b) The wavelength of sound for the stationary observer is 0.4375 m.
(c) The speed of sound for the stationary observer is 350 m/s.

Question 30

Q30. A string fixed at both ends is vibrating in the n-th mode of vibration. Is the amplitude at an antinode equal to the amplitude at any other antinode? Give reasons.

Accepted Answer

Yes, in an ideal string vibrating in a specific mode (n-th mode), the amplitude at an antinode is equal to the amplitude at any other antinode.

**Reason:**
When a string fixed at both ends vibrates in a specific mode, it forms a standing wave. A standing wave is formed by the superposition of two identical progressive waves traveling in opposite directions. Antinodes are points where constructive interference is maximum, resulting in the largest displacement from the equilibrium position. For a given mode of vibration, the conditions for constructive interference are uniform across all antinodal positions along the string, assuming the string is uniform and the medium is ideal. Therefore, all antinodes will exhibit the same maximum amplitude of vibration.

Question 31

Q31. A string fixed at both ends is vibrating in the n-th mode of vibration. Is the energy of vibration of the string in the n-th mode equal to the energy of vibration in the (n+1)-th m

Accepted Answer

No, the energy of vibration of the string in the n-th mode is generally **not** equal to the energy of vibration in the (n+1)-th mode.

**Reason:**
The energy of a vibrating string depends on its amplitude and frequency. For a simple harmonic motion, the energy (E) is proportional to the square of the amplitude (A) and the square of the frequency (f): E ∝ A²f².

For a string fixed at both ends, the frequency of vibration in the n-th mode (f_n) is given by:
f_n = n * (v / 2L)
where 'v' is the speed of the wave on the string and 'L' is the length of the string.

Therefore, the frequency for the (n+1)-th mode will be:
f_(n+1) = (n+1) * (v / 2L)

Clearly, f_n ≠ f_(n+1) since n ≠ (n+1). Specifically, f_(n+1) > f_n.

While the maximum amplitude at the antinodes might be considered the same for different modes if the string is excited with the same initial energy input (though this is often not the case in practice, as higher modes are harder to excite to the same amplitude), the difference in frequency alone means that the total energy of vibration will be different. Since higher modes have higher frequencies, they generally possess higher vibrational energy, assuming comparable amplitudes.

Question 32

Q32. A string fixed at both ends is vibrating in the n-th mode of vibration. Is the frequency of vibration of the string in the n-th mode equal to the frequency of vibration in the (n+1

Accepted Answer

No, the frequency of vibration of the string in the n-th mode is **not** equal to the frequency of vibration in the (n+1)-th mode.

**Reason:**
For a string fixed at both ends, the possible frequencies of vibration (also known as natural frequencies or harmonics) are quantized. The frequency of the n-th mode of vibration (f_n) is given by the formula:

f_n = n * (v / 2L)

Where:
*   `n` is an integer representing the mode number (n = 1 for fundamental mode, n = 2 for first overtone, etc.)
*   `v` is the speed of the transverse wave on the string (v = √(T/μ), where T is tension and μ is linear mass density)
*   `L` is the length of the string

From this formula, it is clear that the frequency is directly proportional to the mode number 'n'.

Therefore:
*   Frequency in the n-th mode: f_n = n * (v / 2L)
*   Frequency in the (n+1)-th mode: f_(n+1) = (n+1) * (v / 2L)

Since n and (n+1) are different integer values, f_n will always be different from f_(n+1). Specifically, f_(n+1) = (n+1)/n * f_n, meaning the frequency increases with higher modes.

Question 33

Q33. A string fixed at both ends is vibrating in the n-th mode of vibration. Is the wavelength of vibration of the string in the n-th mode equal to the wavelength of vibration in the (n

Accepted Answer

No, the wavelength of vibration of the string in the n-th mode is **not** equal to the wavelength of vibration in the (n+1)-th mode.

**Reason:**
For a string fixed at both ends, the possible wavelengths of the standing waves are also quantized. The condition for standing waves requires that the length of the string (L) must be an integral multiple of half-wavelengths.

L = n * (λ_n / 2)

Where:
*   `n` is the mode number (n = 1, 2, 3, ...)
*   `λ_n` is the wavelength of the n-th mode

From this, the wavelength for the n-th mode (λ_n) is given by:

λ_n = 2L / n

From this formula, it is clear that the wavelength is inversely proportional to the mode number 'n'.

Therefore:
*   Wavelength in the n-th mode: λ_n = 2L / n
*   Wavelength in the (n+1)-th mode: λ_(n+1) = 2L / (n+1)

Since n and (n+1) are different integer values, λ_n will always be different from λ_(n+1). Specifically, λ_(n+1) = n / (n+1) * λ_n, meaning the wavelength decreases with higher modes.

Chapter 15: Waves