Question

The speeds of a particle performing linear S.H.M. are 8 cm/s and 6 cm/s at respective displacements of 6 cm and 8 cm. Find its period and amplitude.

Answer 1

Given:

The speeds of a particle performing linear S.H.M. are 8 cm/s and 6 cm/s at respective displacements of 6 cm and 8 cm.

To find:

Time period and amplitude ?

Calculation:

General expression for velocity of a SHM particle:

$\boxed{ \bold{ \therefore \: v = \omega \sqrt{ {A}^{2} - {x}^{2} } }}$

In 1st case:

$\rm \therefore \: v = \omega \sqrt{ {A}^{2} - {x}^{2} }$

$\rm \implies \: 8= \omega \sqrt{ {A}^{2} - {6}^{2} }$

$\rm \implies \: 8= \omega \sqrt{ {A}^{2} - 36 } \: \: \: \: \: \: \: ......(1)$

In 2nd case:

$\rm \therefore \: v = \omega \sqrt{ {A}^{2} - {x}^{2} }$

$\rm \implies \: 6= \omega \sqrt{ {A}^{2} - {8}^{2} }$

$\rm \implies \: 6= \omega \sqrt{ {A}^{2} - 64 } \: \: \: \: \: \: \: .......(2)$

Dividing the equations:

$\therefore \: \dfrac{8}{6} = \dfrac{ \sqrt{ {A}^{2} - 36 } }{ \sqrt{ {A}^{2} - 64} }$

$\implies \: \dfrac{64}{36} = \dfrac{ {A}^{2} - 36 }{ {A}^{2} - 64}$

$\rm\implies \: 64 {A}^{2} - 4096 = 36 {A}^{2} - 1296$

$\rm\implies \:28 {A}^{2} = 2800$

$\rm\implies \: {A}^{2} = 100$

$\rm\implies \: A = 10 \: cm$

So, amplitude is 10 cm.

$\rm \therefore \: 8= \omega \sqrt{ {A}^{2} - 36 } \: \: \: \: \: \: \: ......(1)$

$\rm \implies \: 8= \omega \sqrt{ {(10)}^{2} - 36 }$

$\rm \implies \: 8= \omega \sqrt{64}$

$\rm \implies \: 8= 8\omega$

$\rm \implies \:\omega = 1 \: hz$

So, time period:

$\rm \therefore \: T = \dfrac{2\pi}{ \omega}$

$\rm \implies \: T = \dfrac{2\pi}{1}$

$\rm \implies \: T = 2\pi \: sec$

So, time period is 2π seconds.

Answer 2

answer

T= 6.284 s is the answer hope it's helpful