Question

The time period of a body executing simple harmonic motion is 0.05 sec. If the amplitude is 4 cm then what will be the maximum acceleration of the body??

Answer 1

given that...
T=0.05s
A=4cm
=0.04m

Now,
acceleration, a=Aw*2
and w=2pie/T
put the values and get the answer......

refer to the attachment...

happy helping mate...

Answer 2

The maximum acceleration of the body is $631.65\ m/s^2$.

Explanation:

It is given that,

Time period of a body executing simple harmonic motion is 0.05 sec.

Amplitude of the motion, A = 4 cm = 0.04 m

To find,

The maximum acceleration of the body.

Solution,

The maximum acceleration of the particle that is executing SHM is given by :

$a=A\omega^2$

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

$a=A(\dfrac{2\pi}{T})^2$

$a=0.04\times (\dfrac{2\pi}{0.05})^2$

$a=631.65\ m/s^2$

So, the maximum acceleration of the body is $631.65\ m/s^2$. Hence, this is the required solution.

Learn more,

Simple harmonic motion

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