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. \(1.9~\text{J}\)

Two springs, of force constants k₁ and k₂ are connected to a mass m as shown in the figure. The frequency of oscillation of the mass is f. If both k₁ and k₂ are made four times their original values, the frequency of oscillation will become:
        

A particle is executing SHM according to y = a cos$\mathrm{\omega t}$. Then, which of the following graphs represent variations of potential energy?
             
1. I and III

2. II and IV

3. II and III

4. I and IV

The potential energy of a particle oscillating along the x-axis is given as U = 20+ (x–2)² where U is in joules and x in meters. The total mechanical energy of the particle is 36 J. The maximum kinetic energy of the particle will be:
1. 24 J 

2. 36 J

3. 16 J 

4. 20 J

A simple pendulum of mass m swings about point B between extreme positions A and C. Net force acting on the bob at these three points is correctly shown by:

1.		2.
3.		4.

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 \(K_0\)

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. \( \frac{T}{12}\)

Kinetic energy of a particle executing simple harmonic motion in straight line is \(pv^2\) and potential energy is \(qx^2,\) where \(v\) is speed at distance \(x\) from the mean position. The time period of the SHM is given by the expression:

1. $2\mathrm{\pi }\sqrt{\frac{\mathrm{q}}{\mathrm{p}}}$

2. $2\mathrm{\pi }\sqrt{\frac{p}{q}}$

3. $2\mathrm{\pi }\sqrt{\frac{\mathrm{q}}{\mathrm{p}+\mathrm{q}}}$

4. $2\mathrm{\pi }\sqrt{\frac{p}{p+q}}$

The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it, another mass M is also suspended, the period of oscillation will now be:

1. T

2. T/$\sqrt{2}$

3. 2T

4. $\sqrt{2}$T