Simple hormonic motion

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Question

A simple pendulum is suspended from the roof of a stationary car and the time period of the pendulum is 2 sec. Now the car starts moving with an acceleration which gradually increases from 0 to ‘a’. Now the time period of oscillation is found to be $\sqrt{2} \sec$ . Then the value of ‘a’ in $m / s^{2}$ is

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Solution

When the car is stationary, $T_{1} = 2 π \sqrt{\frac{l}{g}}$
For accelerated car, effective value of ‘g’ is $\sqrt{g^{2} + a^{2}}$
$∴ T_{2} = 2 π \sqrt{\frac{l}{\sqrt{g^{2} + a^{2}}}}$
$∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{\sqrt{g^{2} + a^{2}}}{g}} = a = g \sqrt{{(\frac{T_{1}}{T_{2}})}^{4} - 1} = 10 \sqrt{{(\frac{2}{\sqrt{2}})}^{4} - 1} m / s^{2} = 10 \sqrt{3} m / s^{2}$

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Simple hormonic motion

Remember concepts with our Masterclasses.

A simple pendulum is suspended from the roof of a stationary car and the time period of the pendulum is 2 sec. Now the car starts moving with an acceleration which gradually increases from 0 to ‘a’. Now the time period of oscillation is found to be 2sec. Then the value of ‘a’ in m/s2 is

When the car is stationary, T1=2πlgFor accelerated car, effective value of ‘g’ is g2+a2∴T2=2πlg2+a2∴T1T2=g2+a2g=a=gT1T24-1=10224-1m/s2=103m/s2

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