Questions
The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is
detailed solution
Correct option is A
At mean position, the kinetic energy is maximum. Hence 12ma2ω2=16 12×5.12×25×10-22ω2=16On putting the values we get ω=10 ⇒ T=2πω=π5secTalk to our academic expert!
Similar Questions
A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is . What must be the least period of these oscillations, so that the object is not detached from the platform
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