The equation of motion of a particle is $\frac{d^{2}y}{d{t}^2}+Ky=0$ where K is positive constant. The time period of the motion is given by

(1) $\frac{2\mathrm{\pi }}{K}$             

(2) $2\mathrm{\pi K}$

(3) $\frac{2\mathrm{\pi }}{\sqrt{K}}$           

(4)  $2\pi \sqrt{K}$

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

(1) $\frac{\mathrm{\pi }}{5}sec$       

(2) $2\mathrm{\pi } \sec$

(3) $20\mathrm{\pi } \sec$   

(4) $5\mathrm{\pi } \sec$

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth, then its time period on the other planet will be:

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: 

If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period will become:

1. 3T

2. 3/2T

3. 4T

4. 2T

A simple harmonic wave having an amplitude a and time period T is represented by the equation $y=5 \mathrm{sin\pi }\left(t+4\right)$m Then the value of amplitude (a) in (m) and time period  (T) in second are       

(1)   $a=10, T=2$   

(2) $a=5, T=1$

(3)    $a=10, T=1$    

(4) $a=5, T=2$

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3 then the time period of the pendulum is

(1) $\frac{\mathrm{T}}{\sqrt{3}}$

(2) $\frac{\mathrm{T}}{3}$

(3) $\frac{\sqrt{3}}{2}\mathrm{T}$

(4) $\sqrt{3}\mathrm{T}$

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration $\frac{\mathrm{g}}{3}$ is

(1) $2\mathrm{\pi }\sqrt{\frac{3\mathrm{L}}{\mathrm{g}}}$

(2) $\mathrm{\pi }\sqrt{\left(\frac{3\mathrm{L}}{\mathrm{g}}\right)}$

(3) $2\mathrm{\pi }\sqrt{\left(\frac{3\mathrm{L}}{2\mathrm{g}}\right)}$

(4) $2\mathrm{\pi }\sqrt{\frac{2\mathrm{L}}{3\mathrm{g}}}$

If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

(1) $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{R}}_{\mathrm{e}}}{\mathrm{g}}}$

(2) $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{2{\mathrm{R}}_{\mathrm{e}}}{\mathrm{g}}}$

(3) $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{R}}_{\mathrm{e}}}{2\mathrm{g}}}$

(4) T=2 seconds

If the displacement equation of a particle be represented by $y=A\sin Pt+ B\cos Pt$ , the particle executes

(1)         A uniform circular motion

(2)         A uniform elliptical motion

(3)         A S.H.M.

(4)         A rectilinear motion