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

A mass m is suspended from two springs of spring constant $k_1$ $and$ $k_2$ as shown in the figure below. The time period of vertical oscillations of the mass will be
                

1. $2\mathrm{\pi }\sqrt{\left(\frac{{\mathrm{k}}_1+{\mathrm{k}}_2}{\mathrm{m}}\right)}$

2. $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)}}$

3. $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}\left({\mathrm{k}}_1{\mathrm{k}}_2\right)}{\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)}}$

4. $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)}{\left({\mathrm{k}}_1{\mathrm{k}}_2\right)}}$

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: 

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. \({2a \pi \over T}\)  

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

The angular velocities of three bodies in simple harmonic motion are ${\omega }_1,$ ${\omega }_2,$ ${\omega }_3$ with their respective amplitudes as $A_1,$ $A_2,$ $A_3$. If all the three bodies have the same mass and maximum velocity, then: