A particle is executing simple harmonic motion with frequency \(f\). The frequency at which its kinetic energy changes into potential energy, will be:
1. \(\frac{f}{2}\)
2. \(f\)
3. \(2f\)
4. \(4f\)

There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,$F=-Kx$ where x is the displacement. The total energy of body depends upon -

1.  K, x         

2.  K, a

3.  K, a, x    

4.  K, a, v

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

1.  $\frac{1}{8}E$       

2.  $\frac{1}{4}E$

3.  $\frac{1}{2}E$       

4.  $\frac{2}{3}E$

A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

1. P.E. is maximum when x = 0

2. K.E. is maximum when x = 0

3. T.E. is zero when x = 0

4. K.E. is maximum when x is maximum

­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , then the period of the pendulum will be

1. T

2. $\frac{\mathrm{T}}{4}$

3. $\frac{2\mathrm{T}}{\sqrt{5}}$

4. $2\mathrm{T}\sqrt{5}$

The total energy of a particle, executing simple harmonic motion is

1. $\propto x$                 

2. $\propto x^2$

3. Independent of x 

4.$\propto x^{1/2}$

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}\text{'}}}$,  where $g^{\text{'}}$   is equal to

1. g                                                       

2. g-a

3. g+a

4. $\sqrt{{\mathrm{g}}^2+{\mathrm{a}}^2}$

If the length of second's pendulum is decreased by 2%, how many seconds it will lose per day?

1. 3927 sec

2. 3727 sec

3. 3427 sec

4. 864 sec

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

1. $\sqrt{2\mathrm{gl}(1-\mathrm{sin\theta })}$

2. $\sqrt{2\mathrm{gl}(1+\mathrm{cos\theta })}$

3. $\sqrt{2\mathrm{gl}(1-\mathrm{cos\theta })}$

4. $\sqrt{2\mathrm{gl}(1+\mathrm{sin\theta })}$

A body is executing Simple Harmonic Motion. At a displacement x its potential energy is $E_1$ and at a displacement y its potential energy is $E_2$. The potential energy E at displacement $\left(x+y\right)$ is 

1.  $E=\sqrt{E_1}+\sqrt{E_2}$   

2.  $\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$

3.   $E=E_1+E_2$           

4.  None of these.