# 4 - Simple Harmonic Motion Questions Answers

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.

**Asked By: ABHISHEK KHANDELWAL**

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**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 90^{0}

**Asked By: SARIKA SHARMA**

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**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 m_{1 }and m_{2 }r orbiting about their bodycenter o at distance r_{1} and r_{2} from their body center respectively with a common period T. the squar of time period ?

**Asked By: SHUBHAM VED**

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**Joshi sir comment**

According to the diagram

Gm_{1}m_{2}/(r_{1}+r_{2})^{2} = m_{1}r_{1}ω^{2} = m_{2}r_{2}ω^{2}

compare first part with either second or third part of the given equation for getting ω^{2} then find T^{2}