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A particle is executing a simple harmonic motion. Its maximum acceleration is \alpha and maximum velocity is \beta. Then, its time period of vibration will be :

  • Option 1)

    \frac{\alpha } {\beta }

  • Option 2)

    \frac{{\beta ^2 }} {\alpha }

  • Option 3)

    \frac{{2\pi \beta }} {\alpha }

  • Option 4)

    \frac{{\beta ^2 }} {{\alpha ^2 }}

 

Answers (1)

As we learnt in

Equation of S.H.M. -

a=-frac{d^{2}x}{dt^{2}}= -w^{2}x

w= sqrt{frac{k}{m}}

 

- wherein

x= Asin left ( wt+delta 
ight )

 

 

 \alpha = +w^{2}A    (Maximum acceleration)

\beta = Aw    (Maximum velocity)

w= \frac{\alpha}{\beta}

w= \frac{2 \pi}{T}\ \: \: \: \Rightarrow T=\frac{2 \pi}{w}=\frac{2 \pi \beta}{\alpha}

Correct option is 3.


Option 1)

\frac{\alpha } {\beta }

This option is incorrect.

Option 2)

\frac{{\beta ^2 }} {\alpha }

This option is incorrect.

Option 3)

\frac{{2\pi \beta }} {\alpha }

This option is correct.

Option 4)

\frac{{\beta ^2 }} {{\alpha ^2 }}

This option is incorrect.

Posted by

Vakul

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