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Oscillation (Simple Harmonic Motion) DPP Solutions
Welcome to our detailed solutions for the Oscillation (Simple Harmonic Motion) DPP. Below, you will find step-by-step solutions to each problem, along with explanations to help you understand the concepts better. Use the "Show Answer" buttons to reveal the solutions.
Question 1
Two pendulums differ in lengths by 22 cm. They oscillate at the same place so that one of them makes 30 oscillations and the other makes 36 oscillations during the same time. The lengths (in cm) of the pendulum are :
Options:
- (A) 72 and 50
- (B) 60 and 38
- (C) 50 and 28
- (D) 80 and 58
Solution:
Let the lengths of the two pendulums be \( l_1 \) and \( l_2 \). Given that \( l_1 - l_2 = 22 \) cm.
The time period of a pendulum is given by \( T = 2\pi \sqrt{\frac{l}{g}} \).
Given that one pendulum makes 30 oscillations and the other makes 36 oscillations in the same time, we have:
\[ \frac{T_1}{T_2} = \frac{36}{30} = \frac{6}{5} \]
Since \( T \propto \sqrt{l} \), we have:
\[ \frac{\sqrt{l_1}}{\sqrt{l_2}} = \frac{6}{5} \implies \frac{l_1}{l_2} = \left(\frac{6}{5}\right)^2 = \frac{36}{25} \]
Let \( l_1 = 36k \) and \( l_2 = 25k \). Then:
\[ 36k - 25k = 22 \implies 11k = 22 \implies k = 2 \]
Thus, \( l_1 = 72 \) cm and \( l_2 = 50 \) cm.
Answer: (A) 72 and 50
Question 2
Three waves of the same amplitude have frequencies \( (n - 1) \), \( n \), and \( (n + 1) \) Hz. They superpose on one another to produce beats. The number of beats produced per second is :
Options:
- (A) \( n \)
- (B) 2
- (C) 1
- (D) \( 3n \)
Solution:
When two waves of frequencies \( f_1 \) and \( f_2 \) superpose, the number of beats produced per second is \( |f_1 - f_2| \).
Here, the frequencies are \( n - 1 \), \( n \), and \( n + 1 \) Hz.
The beats produced between \( n - 1 \) and \( n \) is \( |n - (n - 1)| = 1 \) beat per second.
The beats produced between \( n \) and \( n + 1 \) is \( |(n + 1) - n| = 1 \) beat per second.
Thus, the total number of beats produced per second is 1.
Answer: (C) 1
Question 3
A spherical ball of mass \( m_1 \) collides head on with another ball of mass \( m_2 \) at rest. The collision is elastic. The fraction of kinetic energy lost by \( m_1 \) is :
Options:
- (A) \( \frac{4m_1m_2}{(m_1 + m_2)^2} \)
- (B) \( \frac{m_1}{m_1 + m_2} \)
- (C) \( \frac{m_2}{m_1 + m_2} \)
- (D) \( \frac{m_1m_2}{(m_1 + m_2)^2} \)
Solution:
In an elastic collision, the kinetic energy is conserved. The fraction of kinetic energy lost by \( m_1 \) is given by:
\[ \frac{\Delta K}{K} = \frac{4m_1m_2}{(m_1 + m_2)^2} \]
This formula is derived from the conservation of momentum and kinetic energy in elastic collisions.
Answer: (A) \( \frac{4m_1m_2}{(m_1 + m_2)^2} \)
Question 4
Two equal masses are connected by a spring satisfying Hooke’s law and are placed on a frictionless table. The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be \( \omega \). Now one of the masses is stopped. The square of the new angular frequency is :
Options:
- (A) \( \omega^2 \)
- (B) \( \frac{\omega^2}{2} \)
- (C) \( \frac{\omega^2}{3} \)
- (D) \( 2\omega^2 \)
Solution:
Initially, the system consists of two masses connected by a spring. The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{k}{\mu}} \]
where \( \mu = \frac{m_1 m_2}{m_1 + m_2} \) is the reduced mass. For two equal masses \( m \), \( \mu = \frac{m}{2} \).
When one mass is stopped, the system reduces to a single mass \( m \) attached to the spring. The new angular frequency \( \omega' \) is:
\[ \omega' = \sqrt{\frac{k}{m}} \]
Thus, \( \omega'^2 = \frac{k}{m} = \frac{\omega^2}{2} \).
Answer: (B) \( \frac{\omega^2}{2} \)
Question 5
When a compressible wave is sent towards the bottom of the sea from a stationary ship, it is observed that its echo is heard after 2s. If the bulk modulus of elasticity of water is \( 2 \times 10^9 \, \text{N/m}^2 \), the mean temperature of water is \( 4^\circ \)C, and the mean density of water is 1000 kg/m\(^3\), then the depth of the sea will be:
Options:
- (A) 1014 m
- (B) 1414 m
- (C) 2828 m
- (D) 3000 m
Solution:
The speed of sound in water is given by:
\[ v = \sqrt{\frac{B}{\rho}} \]
where \( B \) is the bulk modulus and \( \rho \) is the density of water.
Given \( B = 2 \times 10^9 \, \text{N/m}^2 \) and \( \rho = 1000 \, \text{kg/m}^3 \), we have:
\[ v = \sqrt{\frac{2 \times 10^9}{1000}} = \sqrt{2 \times 10^6} = 1414 \, \text{m/s} \]
The time taken for the echo to return is 2 seconds, so the depth \( d \) of the sea is:
\[ d = \frac{v \times t}{2} = \frac{1414 \times 2}{2} = 1414 \, \text{m} \]
Answer: (B) 1414 m
Question 6
The speed of sound in a mixture of \( n_1 = 2 \) moles of He, \( n_2 = 2 \) moles of H\(_2\) at temperature \( T = \frac{972}{5} \) K is \( n \times 10 \, \text{m/s} \). Find \( n \). (Take \( R = \frac{25}{3} \, \text{J/mole-K} \))
Solution:
The speed of sound in a gas mixture is given by:
\[ v = \sqrt{\frac{\gamma RT}{M}} \]
where \( \gamma \) is the adiabatic index, \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the mixture.
For the mixture of He and H\(_2\):
\[ M = \frac{4 \times 2 + 2 \times 2}{4} = 3 \, \text{g/mol} \]
\[ \gamma = 1 + \frac{2}{f} = 1 + \frac{2 \times (2 + 2)}{2 \times 3 + 2 \times 5} = \frac{3}{2} \]
Thus, the speed of sound is:
\[ v = \sqrt{\frac{3}{2} \times \frac{25}{3} \times \frac{1000}{3} \times \frac{972}{5}} = 900 \, \text{m/s} \]
Therefore, \( n = 90 \).
Answer: 90
Question 7
Match the statements in column-I with the statements in column-II.
| Column-I | Column-II |
|---|---|
| (A) A tight string is fixed at both ends and sustaining standing wave | (p) At the middle, antinode is formed in odd harmonic |
| (B) A tight string is fixed at one end and free at the other end | (q) At the middle, node is formed in even harmonic |
| (C) Standing wave is formed in an open organ pipe. End correction is not negligible. | (r) At the middle, neither node nor antinode is formed |
| (D) Standing wave is formed in a closed organ pipe. End correction is not negligible. | (s) Phase difference between SHMs of any two particles will be either \( \pi \) or zero. |
Solution:
(A) A tight string fixed at both ends will form standing waves with nodes at both ends. In odd harmonics, an antinode is formed at the middle. The phase difference between particles in the same loop is zero, and between adjacent loops is \( \pi \). Thus, (A) matches with (p), (q), and (s).
(B) A tight string fixed at one end and free at the other will not form integral loops, so neither a node nor an antinode is formed at the middle. The phase difference between particles in the same loop is zero, and between adjacent loops is \( \pi \). Thus, (B) matches with (r) and (s).
(C) In an open organ pipe, the standing wave will have antinodes at both ends. The phase difference between particles in the same loop is zero, and between adjacent loops is \( \pi \). Thus, (C) matches with (s).
(D) In a closed organ pipe, the standing wave will have a node at the closed end and an antinode at the open end. The phase difference between particles in the same loop is zero, and between adjacent loops is \( \pi \). Thus, (D) matches with (r) and (s).
Answer: (A) p, q, s; (B) r, s; (C) s; (D) r, s
