A particle executes SHM with time period \(T\). The time period of oscillation of total energy is:
1. \(T\)
2. \(2T\)
3. \(\dfrac{T}{2}\)
4. Infinite

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. \(\dfrac{a\omega}{4}\)
2. \(\dfrac{a\omega}{2\pi}\)
3. \(\dfrac{2a\omega}{\pi}\)
4. \(\dfrac{a\omega}{\sqrt{3}\pi}\)

The time period of oscillation of a simple pendulum of length equal to half of the diameter of the earth is about 

1.  60 minute

2.  84.6 minute

3.  42.3 minute

4.  24 hour

The time period of the given spring-mass system is:

The equation of simple harmonic motion is given by X = (4 cm)$\sin \left(6.28<mspace\; width="1em"></mspace>t<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{5\mathrm{\pi }}{3}\right)$, then maximum velocity of the particle in simple harmonic motion is: 

1.  25.12 m/s 

2.  25.12 cm/s 

3.  12.56 m/s 

4.  12.56 cm/s

A spring pendulum is placed on a rotating table. The initial angular velocity of the table is \(\omega_{0}\) and the time period of the pendulum is \(T_{0}.\) If the the angular velocity of the table becomes \(2\omega_{0},\) then the new time period of the pendulum will be:

If the vertical spring-mass system is dipped in a non-viscous liquid, then:

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:

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.

The time period of a simple pendulum in a stationary trolley is \(T_1.\) If the trolley is moving with a constant speed, then the time period is \(T_2.\) Then: 
1. \(T_1>T _2\)
2. \(T_1<T _2\)
3. \(T_1=T _2\)
4. \(T_2= \infty \)