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: 

1.	\(T_2 ~\text{is infinity} \)	2.	\(\mathrm{T}_2>\mathrm{T}_1 \)
3.	\(\mathrm{T}_2<\mathrm{T}_1 \)	4.	\(T_2=T_1\)

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 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

The displacement of a particle varies according to the relation x = 4(cosp$\mathrm{\pi }$t + sinp$\mathrm{\pi }$t). The amplitude of the particle is

(1) 8

(2) -4

(3) 4

(4) $4\sqrt{2}$

The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\theta$, is given by -

1.   $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gcos\theta }}}$               

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gsin}\theta }}$

3.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g} }}$                     

4. $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gtan\theta }}}$

An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is -

(1) 4 Mg/K         

(2) 2 Mg/K

(3) Mg/K             

(4) Mg/2K

The graph shows the variation of displacement of a particle executing SHM with time. We infer from this graph that: