Question

e) Particle starts from the mean position and executes S.H.M. of period 4s and

amplitude 10cm. Find its displacement at the end of half a second.​

Answer 1

Explanation:

=2s,A=10 cm

Displacement ,x=Asin2πTt

=10sin2π2×16=10sinπ6=5cm

(ii) Velocity , v=dxdt=A2πTcos2πTt

=10×2π2cos2π2×16

=10×3.14×3–√2=27.16 cm/s

(iii) Acceleration , a=dvdt=4π2T2Asin2πTt

4×(3.14)24×10×sin2π2×16

=49.3cm/s2