Question

15. A particle in S.H.M. has a velocity 20
cm/sec, when it crosses the mean
position. Find its velocity when it is
midway between the mean position and
extreme position.​

Answer 1

Answer:

Maximum value of velocity is obtained at the mean position , such that :

$v_{max} = 20 \: cm {s}^{ - 1}$

Now we need to find the velocity when the oscillatory particle passes the half position between the mean and extreme positions.

Velocity at any point is represented as :

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

We know that x = A/2 as per Question :

$= > v = \omega \sqrt{ {A}^{2} - { (\frac{A}{2}) }^{2} }$

$= > v = \omega \sqrt{ {A}^{2} - \dfrac{ {A}^{2} }{4}}$

$= > v = \omega \sqrt{ \dfrac{ {3A}^{2} }{4}}$

$= > v = A\omega (\dfrac{ \sqrt{3} }{2} )$

$= > v = (\dfrac{ \sqrt{3} }{2} ) \times v_{max}$

$= > v = (\dfrac{ \sqrt{3} }{2} ) \times \: 20$

$= > v = 10 \sqrt{3} \: cm {s}^{ - 1}$

So final answer :

$\boxed{ \red{ \bold{ \huge{v = 10 \sqrt{3} \: cm {s}^{ - 1} }}}}$

Answer 2

.....

$\tt{\huge{\purple{ ................ }}}$

QUESTION :

15. A particle in S.H.M. has a velocity 20 cm/sec, when it crosses the mean position.

Find its velocity when it is midway between the mean position and

between the mean position andextreme position.

SOLUTION :

From the above Question, we can gather the following information.....

$\tt{\purple{\leadsto{ v_{0} = \omega_{1} \sqrt{ {A}^ 2 - { x } ^ 2 } }}}$

Here, X = { A / 2 }

Substuting this value into the Equation,

We get :

$\tt{\blue{\leadsto{ V _{0} = \dfrac{\sqrt{3}}{2} A \times v_{max} }}}$

Here, A = 1

Substuting the required values :

$\sf{\orange{\leadsto{ v_{0} = \dfrac{ \sqrt{3} }{2 } \times 1 \times 20 = 10 \sqrt{3} \: m / s. }}}$

ANSWER :

The velocity of the required particle when it is midway between the mean position and extreme position is 10√3 m / s.