# 8 - Simple Harmonic Motion Questions Answers

The quality factor of a sonometer wire is 2 x 10³. On plucking it makes 240 per second. Calculate the time in which the amplitude decreases to half the initial value.

Joshi sir comment
Joshi sir comment

small ring is threaded on an inextensible frictionless cord of length 2l.

The ends of the cord are fixed to a horizontal ceiling. In equilibrium, the

ring is at a depth h below the ceiling. Now the ring is pulled aside by a

small distance in the vertical plane containing the cord and released.

Find period of small oscillations of the ring. Acceleration of free fall is g.

Joshi sir comment

TWO PENDULUM HAVE TIME PERIODS T &5T/4. THEY STARTS SHM AT THE SAME TIME FROM THE MEAN POSITION . WHAT WILL BE THE PHASE DIFFERENCE B/W THEM AFTER D BIGGER PENDULUM COMPLETED ONE OSCILLATn ??

ans 900

Joshi sir comment

When bigger pendulum will complete its one oscillation, the smaller one will complete [1+(1/4)] oscillation. so phase difference = 2π*(T/4)/T = π/2

2 comparable masses m1 and m2  r orbiting about their bodycenter o at distance r1 and r2 from  their  body center respectively  with a common period T. the squar of time period  ?

Joshi sir comment According to the diagram

Gm1m2/(r1+r2)2 = m1r1ω2 = m2r2ω2

compare first part with either second or third part of the given equation for getting ω2 then find T2