Questions
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is
detailed solution
Correct option is C
vmax=aω ⇒ω=vmaxa=104Now, v=ωa2−y2 ⇒v2=ω2(a2−y2) ⇒y2=a2−v2ω2⇒y=a2-v2ω2=42-5210/42=23cmTalk 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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