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.	\(\sqrt{T} \)	2.	\(T \)
3.	\({T}^{1 / 3} \)	4.	\(\sqrt{2} {T}\)

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

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 particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $\mathrm{Fsin\omega t}$ . If the amplitude of the particle is maximum for $\mathrm{\omega }={\mathrm{\omega }}_1$  and the energy of the particle is maximum for $\mathrm{\omega }={\mathrm{\omega }}_2$, then (where ${\omega }_0$ is natural frequency of oscillation of particle)

1. ${\mathrm{\omega }}_1={\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2\ne {\mathrm{\omega }}_0$

2. ${\mathrm{\omega }}_1={\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2={\mathrm{\omega }}_0$

3. ${\mathrm{\omega }}_1\ne {\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2={\mathrm{\omega }}_0$

4. ${\mathrm{\omega }}_1\ne {\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2\ne {\mathrm{\omega }}_0$

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

1.   8           

2.  – 4

3.   4          

4.   $4\sqrt{2}$