Question

The maximum velocity of a particle executing shm is v.If the amplitude is double and the time period of oscillation decreased to 1/3 of its original value.The maximum velocity becomes :

Answer 1

Answer:  The maximum velocity becomes 6 v.

Explanation:

Given that,

Maximum velocity = v

If the amplitude is double and the time period of oscillation decreased to 1/3 of its original value.

We know that,

The maximum velocity is  

$v = a\omega$

The time period of the oscillation is

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

Now, time period of oscillation decreased to 1/3 of its original value

Therefore,

$\omega'=\dfrac{3\times2\pi}{T}$

$\omega'=3\omega$

New amplitude a' = 2 a

The new maximum velocity is

$v' = 2a\times3\omega$

$v' = 6 a\omega$

$v' = 6 v$

Hence, The maximum velocity becomes 6 v.

Answer 2

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The maximum velocity of a particle executing she is v. if the amplitude is double and the time period of oscillation decreased to 1/3 of its original value. the maximum velocity becomes:

$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \large \orange{ \bold{ \underline{ \: step \: by \: step \: explaination}}}$

$\: \\ \implies \displaystyle \sf \: let \: consider \: \: v \: = maximum \: velocity \: of \: a \: particle$

$\: \displaystyle \sf{executing \: harmonic \: motion}.$

$\: \implies \boxed{ \underline{ \bf{v = aw}}}....(1)$

$\: \\ \implies \displaystyle \sf \: after \: the \: time \: decreased \: \frac{1}{3} of \: its \: original \: value$

$\implies\displaystyle\sf{T'=\dfrac{1}{3}T}$

$\implies\displaystyle\sf{A'=2A}$

$\implies\displaystyle\sf{ w'=\dfrac{2π}{T'}=3w}$

$\implies\displaystyle\sf{V'=w'A'}$

$\implies\displaystyle\sf{V'=2A\times 3w}$

$\implies\displaystyle\sf{V'=6aw=6V}$ ......from(1)

$\: \implies \boxed{ \displaystyle \sf { \underline{ \: the \: answer \: will \: be \: 6v}}}$