The motion of the particle is started at t = 0 and the equation of motion is given by $x = 8 \sin \left(100t + \frac{\mathrm{\pi }}{6}\right)$, where x is in cm and t is in seconds. When will the particle come to rest for the first time?

1.  $\frac{\mathrm{\pi }}{300} \mathrm{s}$

2.  $\frac{\mathrm{\pi }}{200} \mathrm{s}$

3.  $\frac{\mathrm{\pi }}{100} \mathrm{s}$

4.  $\frac{\mathrm{\pi }}{400} \mathrm{s}$

What will be the frequency of oscillation of a simple pendulum, if the length of the pendulum is equal to the radius of earth?

1.  $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$

2.  $\frac{1}{\mathrm{\pi }}\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$

3.  $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{2\mathrm{g}}{\mathrm{R}}}$

4.  $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{3\mathrm{g}}{\mathrm{R}}}$

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.

The time period of a body under S.H.M. is ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ when restoring forces ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$ respectively act on it. What will be the time period of S.H.M. when both the forces act simultaneously on it?

1.  $\frac{{\mathrm{T}}_1{\mathrm{T}}_2}{{\mathrm{T}}_1 + {\mathrm{T}}_2}$

2.  ${\mathrm{T}}_1 + {\mathrm{T}}_2$

3.  $\sqrt{\frac{{\mathrm{T}}_1 + {\mathrm{T}}_2}{{\mathrm{T}}_1{\mathrm{T}}_2}}$
4. \(\frac{\mathrm{T_1 T_2}}{\sqrt{\mathrm{T^2_1+T^2_2}}}\)

A particle is subjected to two mutually perpendicular SHM such that x = 2sin$\mathrm{\omega }$t and

$\mathrm{y} = 2 \sin \left[\mathrm{\omega t} + \frac{\mathrm{\pi }}{2}\right]$. The path of the particle will be

1.  An ellipse

2.  A straight line

3.  A parabola

4.  A circle

Suppose that a pendulum clock is carried to a depth of 32 km inside the earth (Radius R = 6400 km). To have the correct time from the clock, by what percentage the effective length of the pendulum should be changed? 

1.  0.5%

2.  -0.5%

3.  1%

4.  -1.0%

When a periodic force \(\vec{F_1}\) acts on a particle, the particle oscillates according to the equation \(x=A\sin\omega t\). Under the effect of another periodic force \(\vec{F_2}\), the particle oscillates according to the equation \(y=B\sin(\omega t+\frac{\pi}{2})\). The amplitude of oscillation when the force (\(\vec{F_1}+\vec{F_2}\)) acts are:

A particle is executing SHM with amplitude A and angular frequency $\mathrm{\omega }$. The time taken by the particle to move from x = 0 to x = A/2 is

1.  $\frac{\mathrm{\pi }}{12 \mathrm{\omega }}$

2.  $\frac{\mathrm{\pi }}{6 \mathrm{\omega }}$

3.  $\frac{\mathrm{\pi }}{3 \mathrm{\omega }}$

4.  $\frac{\mathrm{\pi }}{\mathrm{\omega }}$

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. \(T'=2T\)

A particle executes simple harmonic oscillations under the effect of small damping. If the amplitude of oscillation becomes half of the initial value of 16 mm in five minutes, then what will be the amplitude after fifteen minutes?

1.  8 mm

2.  4 mm

3.  2 mm

4.  1 mm