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Physics - SHM Question with Solution | TestHub

PhysicsSHMComplex situation involving others topicEasy2 minQB

PhysicsEasynumerical

A horizontal platform, with an object placed on it, is executing SHM in the vertical direction. The amplitude of oscillation is $2.5 cm$ . Let the least period of these oscillations, so that the object is not detached, be equal to $\frac{π}{2 a}$ . Then, the value of $a$ is (Given: $g = 10 m s^{- 2}$ )

Answer:

5.00

Solution:

The object would not leave the platform when
$a_{m a x} < g$
$∴ w^{2} A < g$
$∴ w < \sqrt{\frac{10}{2.5 \times 10^{- 2}}} = \sqrt{\frac{10^{4}}{25}} = \frac{100}{5} = 20$
$∴ \frac{2 π}{T} < 20 \Rightarrow T > \frac{π}{10}$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Complex situation involving others topic

⏱ 2mℹ️ Source: QB

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