# A spring pendulum is on the rotating table. The initial angular velocity of the table is $$\omega_{0}$$ and the time period of the pendulum is $$T_{0}.$$ Now the angular velocity of the table becomes $$2\omega_{0},$$ then the new time period will be: 1. $$2T_{0}$$ 2. $$T_0\sqrt{2}$$ 3. remains the same 4. $$\frac{T_0}{\sqrt{2}}$$

Subtopic:  Angular SHM |
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The displacement $$x$$ of a particle varies with time $$t$$ as $$x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)$$$\mathrm{}$The time taken by the particle to reach from $$x = \frac{A}{2}$$ to $$x = -\frac{A}{2}$$ will be:

 1 $$\frac{T}{2}$$ 2 $$\frac{T}{3}$$ 3 $$\frac{T}{12}$$ 4 $$\frac{T}{6}$$

Subtopic:  Phasor Diagram |
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Force on a particle F varies with time t as shown in the given graph. The displacement x vs time t graph corresponding to the force-time graph will be:

 1 2 3 4
Subtopic:  Linear SHM |
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A particle executes SHM with a frequency of $$20$$ Hz. The frequency with which its potential energy oscillates is:
1. $$5$$ Hz
2. $$20$$ Hz
3. $$10$$ Hz
4. $$40$$ Hz

Subtopic:  Energy of SHM |
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The curve between the potential energy $$(U)$$ and displacement $$(x)$$ is shown. Which of the oscillation is about the mean position, $$x = 0?$$

 1 2 3 4
Subtopic:  Energy of SHM |
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A spring-block system oscillates with a time period $$T$$ on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes $$T'.$$ Then one can conclude that:
1. $$T'>T$$
2. $$T'<T$$
3. $$T'=T$$
4. $$T'=2T$$

Subtopic:  Combination of Springs |
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A particle executes linear SHM between $$x=A.$$ The time taken to go from $$0$$ to $$A/2$$ is ${\mathrm{T}}_{1}$ and to go from $$A/2$$ to $$A$$ is ${\mathrm{T}}_{2}$, then:

 1 ${\mathrm{T}}_{1}$ $<$ 2 ${\mathrm{T}}_{1}$ $>$ ${\mathrm{T}}_{2}$ 3 ${\mathrm{T}}_{1}$ $=$ ${\mathrm{T}}_{2}$ 4 ${\mathrm{T}}_{1}$ $=$ $2{\mathrm{T}}_{2}$
Subtopic:  Linear SHM |
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Two simple pendulums of length 1 m and 16 m are in the same phase at the mean position at any instant. If T is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:

1.  $\frac{3\mathrm{T}}{2}$

2.  $\frac{3\mathrm{T}}{4}$

3.  $\frac{2\mathrm{T}}{3}$

4.  $\frac{4\mathrm{T}}{3}$

Subtopic:  Angular SHM |
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A particle executes SHM with a time period of 4 s. The time taken by the particle to go directly from its mean position to half of its amplitude will be:

1.  $\frac{1}{3}$ s

2.  1 s

3.  $\frac{1}{2}$ s

4.   2 s

Subtopic:  Linear SHM |
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The graph between the velocity (v) of a particle executing S.H.M. and its displacement (x) is shown in the figure. The time period of oscillation for this SHM will be

1.  $\sqrt{\frac{\mathrm{\alpha }}{\mathrm{\beta }}}$

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{\alpha }}{\mathrm{\beta }}}$

3.  $2\mathrm{\pi }\left(\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)$

4.  $2\mathrm{\pi }\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}\right)$

Subtopic:  Simple Harmonic Motion |
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