Test Index

Oscillations Part 1

Section: Physics

Question : 63 of 100

Marks: +1, -0

Solution:

Time taken to reach the extreme position from equilibrium position is

\frac{T}{4}

.Velocity is maximum at equilibrium position and zero at extreme position.

V = A\omega\cos\omega t

K.E

=\frac{1}{2}mv^2

(m is the mass of particle and v is the velocity of particle)

K.E.

=\frac{1}{2} m A^2\omega^2\cos^2\omega t

Hence graph of K.E. v/s time is square cos function

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