Question

A particle executing simple harmonic motion has amplitude of 1m and time period pie sec.Velocity of particle when displacement is 0.8m is​

Answer 1

Answer:1.2m/s

Explanation:

V=w√A^2-y^2

Where w-angular velocity

A is amplitude

Y is displacement

W=2π÷T

T =10

So,w=2

V=2√1^2-0.8^2

=2√0.36

=2*0.6

=1.2

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