In a simple pendulum, the period of oscillation \(T\) is related to the length of the pendulum \(L\) as:
1. \(\frac{L}{T}= \text{constant}\)
2. \(\frac{L^2}{T}= \text{constant}\)
3. \(\frac{L}{T^2}= \text{constant}\)
4. \(\frac{L^2}{T^2}= \text{constant}\)

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 particle in SHM is described by the displacement equation $\mathrm{x}\left(\mathrm{t}\right)=\mathrm{Acos} \left(\mathrm{\omega t}+\mathrm{\theta }\right).$ If the initial position of the particle is 1 cm and its initial velocity is $\mathrm{\pi }$cm/s, what is its amplitude? (The angular frequency of the particle is $\pi s^{-1}$)

(1)   1 cm       

(2)  $\sqrt{2}$ cm

(3)    2 cm     

(4)   2.5 cm

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. \(\dfrac{3}{2}T\)
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) $2\mathrm{\pi }\sqrt{\frac{2\mathrm{L}}{3\mathrm{g}}}$