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1.

A pendulum  of length 1m  is released  from theta =60 degree . The rate of change of speed  of the bob at  theta = 30 degree is

Answer»

We know that Rate change of speed, dv/dt= tangential acceleration =tangential force/mass

When θ=30∘, tangential force is mgsinθ

Therefore F=mgsin30

ma=mgsin30

 a=mgsin30/m

=10sin30

=10×1/2

=5m/s2

So, The rate of change of speed  of the bob at  theta = 30 degree is 5m/s2
Discussion
2.

The equation  of shm  y= asin (2pi Nt  + â) . Calculate  the phase of the given equation  at time =t

Answer» Phase difference  is â
Discussion
3.

The two particle  are oscillating along two close  parallel straight line in opposite  direction towards the mean position The time taken by the particle  X=0 to x=A/2 is T/12 and x=A to x=A/2 is T/6. Then calculate the time difference.Ans-T/3

Answer» Time taken by the  first particle is   t/12 and second one is t/6 but first one has to reaches the mid point   ( that point they cross each other)  therefore  first particle will     crossed  before  attaining  the t/12 time .   ( both meet at t/6 time. )That's the REASON  why  it is  not considered   t/12 and used the value t/6.

Both  moving in opposite  direction , so   one particle  should  be-t/6)

Next,  time difference t/6-(-t/6) =t/3
Discussion