Question

In the SHM, the ratio of frequency of oscillation KE to that of velocity v, is​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

In the SHM, the ratio of frequency of oscillation to K.E that of velocity v

• 1 : 2

• 2 : 1

• 4 : 1

• 1 : 4

EVALUATION

Let the equation of SHM be

$\displaystyle \sf{x = a \: \sin \omega t} \: \: \: - - - - (1)$

Frequency of the oscillations $\displaystyle \sf{ = \omega }$

Differentiating both sides of Equation 1 with respect to t we get

$\displaystyle \sf{ \frac{dx}{dt} = \frac{d}{dt} \big( a \: \sin \omega t \big)}$

$\displaystyle \sf{ \implies \: v = \frac{dx}{dt} = a \omega \cos \omega t }$

Frequency of velocity $\sf{ = \omega}$

Now Kinetic energy = K.E

$\displaystyle \sf{ = \frac{1}{2}m {v}^{2} }$

$\displaystyle \sf{ = \frac{1}{2}m {(a \omega \cos \omega t)}^{2} }$

$\displaystyle \sf{ = \frac{1}{2}m {a}^{2} { \omega }^{2} {\cos }^{2}\omega t }$

$\displaystyle \sf{ = \frac{1}{4}m {a}^{2} { \omega }^{2} \times 2 {\cos }^{2}\omega t }$

$\displaystyle \sf{ = \frac{1}{4}m {a}^{2} { \omega }^{2} (1 + {\cos }^{}2\omega t )}$

So frequency of K.E $\displaystyle \sf{ = 2\omega }$

Hence the required ratio

Hence the required ratio

$\sf{ = 2 \omega : \omega}$

$\sf{ = 2 :1 }$

FINAL ANSWER

Hence the correct option is 2 : 1

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