Question

A particle performing linear S.H.M. has
maximum velocity of 20 cm/s and maximum
acceleration of 80 cm/s2.Find the amplitude and
period of oscillation.​

Answer 1

Answer:

Given :

Max Velocity is 20 cm/s

Max acceleration = 80 cm/s²

To find:

Amplitude and time period

Calculation:

Max Velocity = ωA

=> ωA = 20 cm/s .......eq(1)

Max Acceleration = (ω²)A

=> (ω²)A = 80 cm/s² .........eq.(2)

Dividing eq(2) by eq(1)

=> ω = 4 Hz

Time period = 2π/ω

=> Time period = 2π/4

=> Time period = π/2 seconds

Amplitude = 20/ω .....from eq.(1)

=> Amplitude = 20/(4)

=> Amplitude = 5 cm.

So final answer :

$amplitude = 5 \: cm \\ time \: period = \dfrac{\pi}{2} seconds$

Answer 2

$\huge{\underline{\underline{\sf{\pink{Answer:-}}}}}$

${\underline{\underline{\color{darkblue}\bf{Given:-}}}}$

•Max velocity= 20 cm/s

•Max acceleration= 80 cm/s²

${\underline{\underline{\color{darkblue}\bf{To\: Find:-}}}}$

• Amplitude=??

• Time period=??

${\underline{\underline{\color{darkblue}\bf{Solution:-}}}}$

ATQ,

$\sf{\omega\:A=20 ......(1)}$

$\sf{\omega^{2}\:A=80.....(2)}$

Dividing (2) by (1) ,

$\sf{{ \frac{\omega { \cancel{{}^{2} }} { \cancel{\:A }}}{ { \cancel{\omega}}\:{ \cancel{ A} }}}= \frac{8 {\cancel{0}}}{2{ \cancel{0} }}} \\ \\ { \sf{\omega = 4}}$

Frequency= 4 Hz

Now, Time period=$</p><p>\huge{\sf{\frac{2\pi}{\omega}}}$

$\sf{\frac{ \cancel{2}\pi}{ \cancel4}} \\ \\ \: { \boxed{\sf{time \: period = \frac{\pi}{2} \: s}}}$

A=$\huge{\sf{\frac{20}{4}}}$....from(1)

A = 5 cm

$\boxed{\sf{amplitude = 5\:cm}}$