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

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)   $4\sqrt{2}$