Oscillations and Waves Questions and Answers

Answer: C

The pendulum completes 32 oscillations.

Frequency is the number of oscillations per second, so a frequency of 2 Hz means 2 oscillations every second:

\[ 2\ \text{s}^{-1} \times 16\ \text{s} = 32 \]

8 is what you get by dividing 16 by 2, as if 2 Hz were the time period of 2 seconds. 16 ignores the frequency and 64 doubles the correct answer.

Answer: B

The correct statement is that its time period is 0.1 s and frequency is 10 Hz.

\[ \begin{aligned} T &= \frac{3\ \text{s}}{30} = 0.1\ \text{s} \\ f &= \frac{30}{3\ \text{s}} = 10\ \text{Hz} \end{aligned} \]

The time period is the time for one oscillation and the frequency is the number of oscillations per second, so \(f = 1/T\). The options with 10 s and 0.3 Hz invert or confuse the two quantities, and 0.3 s with 0.3 Hz cannot both be right, since their product must be 1.

Answer: B

The correct statement is that the time period calculated by Seema is greater than that calculated by Ashutosh.

In one complete oscillation the bob goes from one extreme to the other and back, covering \(2D\). So each observation gives the period \(T\) as the time per \(2D\).

\[ \begin{aligned} \text{Seema: } T &= 2 \times \frac{2D}{D/4} = 2 \times 8 = 16\ \text{s} \\ \text{Ashutosh: } T &= 10 \times \frac{2D}{3D/2} \approx 13.3\ \text{s} \end{aligned} \]

So Seema's value, 16 s, is greater. Neither is 12 s. The last option is wrong on physics as well: the period of a simple pendulum depends on its length and g, not on the mass of the bob.

Answer: C

X, Y and Z are guitar, tabla, shehnai.

A guitar produces sound from vibrating stretched strings, a tabla from its stretched membrane (the drum skin), and a shehnai, a wind instrument, from its vibrating column of air.

The other sets break the order. In option A the harmonium is a reed and air instrument and the dholak a membrane, both in the wrong places. Option B starts with the dholak, a membrane, where strings are needed. In option D the violin and piano are both string instruments, so neither can fill the membrane or air-column blank.

Answer: A

Both statements are true and the Reason is the correct explanation of the Assertion.

The ceilings of cinema and concert halls are curved so that sound striking them is reflected and spread to all parts of the hall, so that listeners everywhere hear evenly. This is the textbook application of the reflection of sound (NCERT class IX).

The Reason, that sound reflected from a curved surface spreads out in all directions, is the principle behind the design, so it explains the Assertion. The options treating either statement as false, or the link as unrelated, are therefore wrong.

Answer: A

In simple harmonic motion the maximum speed is at the equilibrium position and equals the amplitude times the angular frequency.

\[ \begin{aligned} \omega = \sqrt{\frac{k}{m}} \\ &= \sqrt{\frac{500}{5}} = 10\ \text{rad/s} \end{aligned} \]

\[ v_{max} = A\omega = 0.10 \times 10 = 1\ \text{m/s} \]

Energy gives the same result: \(\tfrac12 kA^2 = \tfrac12 mv^2\) gives \(v^2 = 500 \times 0.01 / 5 = 1\). The amplitude is the 10 cm the block is displaced and released from, which must be changed to metres. The answer is 1 m/s, option A.

Answer: C

The speed of a transverse wave on a stretched string is \(v = \sqrt{T/\mu}\), where \(\mu\) is the mass per unit length.

\[ \begin{aligned} \mu = \frac{4.8 \times 10^{-3}}{1.2} \\ &= 4 \times 10^{-3}\ \text{kg/m} \end{aligned} \]

\[ \begin{aligned} v = \sqrt{\frac{10}{4 \times 10^{-3}}} \\ &= \sqrt{2500} = 50\ \text{m/s} \end{aligned} \]

The other speeds would need a tension of 3.6 N, 6.4 N or 14.4 N respectively for this wire, so only 50 m/s fits 10 N. The answer is option C.

Answer: D

The correct match is P-5, Q-3, R-4, S-1.

This is the force-voltage analogy, where the mechanical equation \(m\ddot{x} + b\dot{x} + kx = F\) is matched term by term with the series circuit equation \(L\ddot{q} + R\dot{q} + \frac{q}{C} = V\):

• mass \(m\) corresponds to inductance \(L\);
• spring constant \(k\) corresponds to reciprocal capacitance \(\frac{1}{C}\);
• displacement \(x\) corresponds to charge \(q\);
• velocity \(\dot{x}\) corresponds to current \(\dot{q}\).

The options with S-2 pair velocity with voltage; voltage corresponds to force. The options with P-3, Q-5 swap mass and spring constant.

Answer: B

The answer is \(f_{A} = f_{B} \gt f_{s}\).

The Doppler shift depends on the component of the source's velocity along the line joining source and observer. Both vehicles move straight towards O, so for each the whole speed lies along that line:

\[ \begin{aligned} f &= f_{s}\,\frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{source}}} \\ &\gt f_{s} \end{aligned} \]

The speeds are equal, so the two heard frequencies are equal. The angle \(\theta\) between the two paths does not matter, because neither vehicle is moving at an angle to its own line of sight. Since both approach, both frequencies are raised, ruling out the options with \(f \lt f_{s}\).

Answer: C

The group velocity is \(\frac{\Delta\omega}{\Delta k}\).

Adding the two waves gives a carrier at the mean frequency, modulated by an envelope:

\[ \begin{aligned} y &= 2a\cos\left(\frac{\Delta\omega\, t - \Delta k\, z}{2}\right) \\ &\quad \times \cos(\bar{\omega} t - \bar{k} z) \end{aligned} \]

The envelope, which carries the energy, moves with speed \(\frac{\Delta\omega/2}{\Delta k/2} = \frac{\Delta\omega}{\Delta k}\), which is the group velocity \(d\omega/dk\) in the limit of small differences.

The options built from \(\omega \pm \Delta\omega\) over \(k \pm \Delta k\) are phase velocities, of one wave or of the carrier, not the speed of the envelope.