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

(1) $\sqrt{\mathrm{T}}$

(2) T

(3) ${\mathrm{T}}^{1/3}$

(4) $\sqrt{2}$T

A particle in SHM is described by the displacement equation $x\left(t\right)=A\cos \left(\omega t+\theta \right). If the initial$ 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

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, the time period is T₂, then

(1) ${\mathrm{T}}_2$ is infinity

(2) ${\mathrm{T}}_2>{\mathrm{T}}_1$

(3) ${\mathrm{T}}_2<{\mathrm{T}}_1$

(4) ${\mathrm{T}}_2={\mathrm{T}}_1$

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 becomes

(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

(a) $\frac{\mathrm{T}}{\sqrt{3}}$

(b) $\frac{\mathrm{T}}{3}$

(c) $\frac{\sqrt{3}}{2}\mathrm{T}$

(d) $\sqrt{3}\mathrm{T}$

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

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

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

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

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

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

A S.H.M. is represented by $x=5\sqrt{2}\left(\sin 2\mathrm{\pi t}+\cos 2\mathrm{\pi t}\right).$ The amplitude of the S.H.M. is

(1)   10 cm         

(2)  20 cm

(3)   $5\sqrt{2}$ cm     

(4)  50 cm