Q. No.If a particle in SHM has a time period of \(0.1\) s and an amplitude of \(6\) cm, then its maximum velocity will be:
1. \(120 \pi\) cm/s
2. \(0.6 \pi\) cm/s
3. \(\pi\) cm/s
4. \(6\) cm/s
1. \(120 \pi\) cm/s
2. \(0.6 \pi\) cm/s
3. \(\pi\) cm/s
4. \(6\) cm/s
| 1. | Zero | 2. | \(30~\text{J}\) |
| 3. | \(20~\text{J}\) | 4. | \(40~\text{J}\) |
1. \(\frac{\pi}{2}~\text{s}\)
2. \(\frac{1}{2}~\text{s}\)
3. \(\pi~\text{s}\)
4. \(1~\text{s}\)
| 1. | \(3~\text{cm}\) | 2. | \(3.5~\text{cm}\) |
| 3. | \(4~\text{cm}\) | 4. | \(5~\text{cm}\) |
1. \(1:\sqrt{3}\)
2. \(1:1\)
3. \(2:1\)
4. \(\sqrt{3}:2\)
1. \(\pi\)
2. \(\frac{\pi}{2}\)
3. \(\frac{\pi}{4}\)
4. \(2\pi\)
A particle is executing linear simple harmonic motion with an amplitude \(a\) and an angular frequency \(\omega\). Its average speed for its motion from extreme to mean position will be:
1. \(\frac{a\omega}{4}\)
2. \(\frac{a\omega}{2\pi}\)
3. \(\frac{2a\omega}{\pi}\)
4. \(\frac{a\omega}{\sqrt{3}\pi}\)
1. \(\frac{1}{2}m\omega^2a^2\)
2. \(\frac{5}{4}m\omega^2a^2\)
3. \(\frac{3}{2}m\omega^2a^2\)
4. Zero
All the surfaces are smooth and springs are ideal. If a block of mass \(m\) is given the velocity \(v_0\) in the right direction, then the time period of the block shown in the figure will be:
1. \(\frac{12l}{v_0}\)
2. \(\frac{2l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
3. \(\frac{4l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
4. \( \frac{\pi}{2}\sqrt{\frac{m}{k}}\)
In a spring pendulum, in place of mass, a liquid is used. If liquid leaks out continuously, then the time period of the spring pendulum:
| 1. | Decreases continuously |
| 2. | Increases continuously |
| 3. | First increases and then decreases |
| 4. | First decreases and then increases |
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