An SHM has an amplitude \(a\) and  a time period \(T.\) The maximum velocity will be:
1. \({4a \over T}\)       
2.  \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)  

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, then the time period becomes T₂. It can be concluded that: 

Two simple harmonic motions of angular frequency 100 rad s ^-1 and 1000 rad $s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. 1:10
2. 1:10²
3. 1:10³                               
4. 1:10⁴

If the displacement x and the velocity v of a particle executing simple harmonic motion are related through the expression $4v^2=25-x^2$,then its time period will be:

A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them will be:

1.  $\frac{\mathrm{\pi }}{3}$

2.  $\frac{2\mathrm{\pi }}{3}$

3.  $\frac{\mathrm{\pi }}{6}$

4.  $\frac{\mathrm{\pi }}{2}$

A point performs simple harmonic oscillation of period \(\mathrm{T}\) and the equation of motion is given by; \(x=a \sin (\omega t+\pi / 6)\). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. \( \frac{T}{8} \)

2. \( \frac{T}{6} \)

3. \(\frac{T}{3} \)

4. \( \frac{T}{12}\)

A particle executing simple harmonic motion has a kinetic energy of \(K_0 cos^2(\omega t)\). The values of the maximum potential energy and the total energy are, respectively:
1. \(0\) and \(2K_0\)
2. \(\frac{K_0}{2}\) and \(K_0\)
3. \(K_0\) and \(2K_0\)
4. \(K_0\) and \(K_0\)

The radius of the circle, the period of revolution, initial position and direction of revolution are indicated in the figure.

The \(y\)-projection of the radius vector of rotating particle \(P\) will be:
1. \(y(t)=3 \cos \left(\frac{\pi \mathrm{t}}{2}\right)\), where \(y\) in m
2. \(y(t)=-3 \cos 2 \pi t\) , where \(y\) in m
3. \(y(t)=4 \sin \left(\frac{\pi t}{2}\right)\), where \(y\) in m
4. \(y(t)=3 \cos \left(\frac{3 \pi \mathrm{t}}{2}\right) \),  where \(y\) in m

A block of mass \(4~\text{kg}\) hangs from a spring of spring constant \(k = 400~\text{N/m}\). The block is pulled down through \(15~\text{cm}\) below the equilibrium position and released. What is its kinetic energy when the block is \(10~\text{cm}\) below the equilibrium position? [Ignore gravity]
1. \(5~\text{J}\)
2. \(2.5~\text{J}\)
3. \(1~\text{J}\)
4. \(1.9~\text{J}\)

The amplitude and the time period in an S.H.M. are 0.5 cm and 0.4 sec respectively. If the initial phase is $\mathrm{\pi }/2$ radian, then the equation of S.H.M. will be:

1. $\mathrm{y}=0.5\sin 5\mathrm{\pi t}$

2. $\mathrm{y}=0.5\sin 4\mathrm{\pi t}$

3. $\mathrm{y}=0.5\sin 2.5\mathrm{\pi t}$

4. $\mathrm{y}=0.5\cos 5\mathrm{\pi t}$