Question

The displacement time graph of a particle executing (S.H.M.) is shown in figure. Whether the following statement is true or not?



A) The force is zero at $t=\frac{3T}{4}$

B) The acceleration is maximum at $t=\frac{4T}{4}.$

C) The velocity is maximum at $t=\frac{T}{4}.$

D) The P.E. is equal to K.E of oscillation at $t=\frac{T}{2}$

Answer 1

A) Formula Used: $F=ma$

Let displacement of the particle at any time t,

$y=A_0\cos \omega t=A_0\cos \frac{2\pi }{T}t\because (\omega =\frac{2\pi }{t})$

So, the acceleration will be $a=-{\omega }^{2}y$

$a=-{\omega }^2{A}_0\cos \frac{2\pi }{T}t$

Force $F=m{\omega }^{2}y.$

At $t=\frac{3T}{4},$ $y=A_0\cos (\frac{2\pi }{T}\times \frac{3T}{4})$

$=A_0\cos \left(\frac{3\pi }{2}\right)=0$

Hence $F=0.$

Final Answer: True


B) Let displacement of the particle at any time $t,$

$y=A_0\cos \omega t=A_0\cos \frac{2\pi }{T}t$

So, the acceleration will be $A=-{\omega }^{2}y$

$a=-{\omega }^2{A}_0\cos \frac{2\pi }{T}t.$

At$t=\frac{4T}{4},$ $y=A_0\cos (\frac{2\pi }{T}\times T)$

$=A_0\cos 2\pi =A_0$

$\therefore \left|a\right|={\omega }^2{A}_0,$ which is maximum value of a.

Final Answer: True.


C) Formula Used: $v=\omega \sqrt{A_0^2-y^2}$

Displacement of the particle at any time $t,$

$y=A_0\cos \omega t=A_0\cos \frac{2\pi }{T}t$

As the velocity is given by

$v=\omega \sqrt{A_0^2-y^2}$

At $t=\frac{T}{4}$

$y=A_0\cos (\frac{2\pi }{T}\times \frac{T}{4})=A_0\cos \left(\frac{\pi }{2}\right)$

$y=0$

Hence, $v=\omega A_0,$ which is maximum value of velocity.

Final Answer: True.


D) Displacement of the particle at any time $t,$

$y=A_0\cos \omega t=A_0\cos \frac{2\pi }{T}t$

At $t=\frac{T}{2},$

$y=A_0\cos (\frac{2\pi }{T}\times \frac{T}{2})=-A_0$

Displacement is maximum, i.e. corresponds to extreme position, it means $PE$ is maximum and $KE$ is zero.

Final Answer: Not true