First slide

Simple hormonic motion

Question

An object of mass 0.2 kg executes simple harmonic along X-axis with frequency of $\frac{25}{π}$ Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J amplitude of oscillation in meter is equal to

Moderate

A

0.05
B

0.06
C

0.01
D

None of the these

Solution

$E = \frac{1}{2} {mω}^{2} A^{2} \Rightarrow E = \frac{1}{2} m {(2 πf)}^{2} A^{2}$
$\Rightarrow A = \frac{1}{2 πf} \sqrt{\frac{2 E}{m}}$
Putting values we obtain,
$A = \frac{1}{2 π (\frac{25}{π})} \sqrt{\frac{2 \times (0.5 + 0.4)}{0.2}} \Rightarrow A = 0.06 m$

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