Question

A particle executing linear SHM has a maximum velocity 40 cm per second and maximum acceleration 50 cm per second square find ots time period​

Answer 1

the answer is given in pics hope it helps you...

Answer 2

The time period of the particle that is executing SHM is 5.02 seconds.

Explanation:

It is given that,

The maximum velocity of the particle executing SHM is, $v_{max}=40\ cm/s=0.4\ m/s$

The maximum acceleration of the particle executing SHM is, $a_{max}=50\ cm/s=0.5\ m/s$

We need to find the time period of the particle. The maximum velocity and acceleration of the particle that is executing SHM is given by :

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

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

Dividing equation (1) and (2). So,

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

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

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

$T=2\pi \times \dfrac{0.4}{0.5}$

T = 5.02 seconds

So, the time period of the particle that is executing SHM is 5.02 seconds. Hence, this is the required solution.

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