Question

If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude the time period will be
1.57 s
3.14 s
6.28 s
12.56 s

Answer 1

Answer:

6.28

Explanation:

I didn't get the question whether it is saying

Vmax = Amax OR Vmax = A

considering the first case

w =1 so T = 6.28

Answer 2

The time period is 6.28 sec.

(c) is correct option.

Explanation:

We know that,

Maximum velocity $v_{max}=\omega^2 A$

Maximum acceleration $a_{max}=\omega A$

We need to calculate the angular velocity

If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude

$v_{max}=a_{max}$

Put the value into the formula

$\omega^2 A=\omega A$

$\omega=1$

We need to calculate the time period

Using formula of time period

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

Put the value into the formula

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

$T=2\times3.14$

$T=6.28\ sec$

Hence, The time period is 6.28 sec.

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Topic : Time period

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