Question

The maximum speed of S.H.M. is 1 m/s and its maximum acceleration is particle executing 1.57m/sec. The time period of the particle will be​

Answer 1

Answer:

T=4 sec

Explanation:

maximum velocity of s.h.m=Aw,

maximum acceleration=aw2,

putting value and solving,we get w=1.57,

and

T=2pi/w

Answer 2

The time period of the particle will be​ 4 seconds.

Explanation:

It is given that,

The maximum speed of the particle in SHM, $v_{max}=1\ m/s$

The maximum acceleration of the particle in SHM, $a_{max}=1.57\ m/s^2$

To find,

The time period of the particle.

Solution,

The maximum speed and acceleration of the particle in SHM is given by :

$v_{max}=A\omega$..............(1)

$a_{max}=A\omega^2$................(2)

$\dfrac{v_{max}}{a_{max}}=\dfrac{1}{\omega}$

Since, $\omega=\dfrac{2\pi}{T}$

$T=2\pi \dfrac{v_{max}}{a_{max}}$

$T=2\pi \dfrac{1}{1.57}$

T = 4 seconds

So, the time period of the particle will be​ 4 seconds.

Learn more

Simple Harmonic motion

https://brainly.in/question/14468217