Question

The maximum displacement of the particle executing shm is 1cm and maximum acceleration is (1.57)^2 its time period is

Answer 1

Answer:

Maximum acceleration= Aomega2omega2

⇒(π2)2=1(2πT)2 (∵ 1.57=π2) ⇒(π2)2=1(2πT)2 (∵ 1.57=π2)

⇒π24=4π2T2 ⇒π24=4π2T2

⇒T2=4×4π2π2 ⇒ T2=4×4π2π2

⇒T2=16 ⇒ T2=16

⇒ T=4sec

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Answer 2

The time period is 0.04 sec.

Explanation:

Given that,

Maximum displacement = 1 cm

Maximum acceleration = 1.57 m/s²

We know that,

The maximum velocity

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

The maximum acceleration is

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

Divide equation (II) by equation (I)

$\dfrac{a}{v}=\dfrac{\omega^2 A}{\omega A}$

$\omega=\dfrac{a}{v}$

Put the value into the formula

$\omega=\dfrac{1.57}{1\times10^{-2}}$

$\omega=157\ rad/s$

We need to calculate the time period

Using formula of angular velocity

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

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

Put the value into the formula

$T=\dfrac{2\pi}{157}$

$T=0.04\ sec$

Hence, The time period is 0.04 sec.

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Topic :

https://brainly.in/question/14468217