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 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. \(y(t)=3 \cos \left(\dfrac{\pi \mathrm{t}}{2}\right)\), where \(y\) in m
2. \(y(t)=-3 \cos 2 \pi t\) , where \(y\) in m
3. \(y(t)=4 \sin \left(\dfrac{\pi t}{2}\right)\), where \(y\) in m
4. \(y(t)=3 \cos \left(\dfrac{3 \pi \mathrm{t}}{2}\right) \),  where \(y\) in m

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