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Physics - SHM Question with Solution | TestHub

PhysicsSHMEnergyEasy2 minQB

PhysicsEasynumerical

The amplitude of a simple pendulum is 10cm. When the pendulum is at a displacement of 4cm from the mean position, then what is the ratio of kinetic and potential energies at that point?

Answer:

5.25

Solution:

$\frac{K . E}{P . E} = \frac{\frac{1}{2} m w^{2} (A^{2} - x^{2})}{\frac{1}{2} m w^{2} x^{2}}$

$\frac{K . E}{P . E} = \frac{A^{2} - x^{2}}{x^{2}}$

$= \frac{10^{2} - 4^{2}}{4^{2}}$

$= \frac{84}{16}$

$\frac{K . E}{P . E} = \frac{21}{4} = 5.25$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: QB

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