Q1 2 particles execute SHM of same amplitude and frequncy on parallel lines. They pass one another when moving in opposite directions each time their displacement is half the amplitude. What is the phase diffrence between them.
ANSWER:2\tiny \prod{} /3
Q2 The length of a spring is \alpha when a force of 4 N is applied on it. The length is \beta when5N is applied, then the length of spring when 9N is applied is
ANSWER:5\beta- 4\alpha
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2 Answers
492)
sprg const=k
5=-kβ, β=-5k
4=-kα, α=-4k
9=-kγ
γ=-9k=-25-16k=-5X5k-(-4X4k)=5β-4α
49A2=Asin(ωt) = Asin(ωt+α)
v=Aωcos(ωt)=-Aωcos(ωt+α)
gives, sin(ωt+α)=sin(ωt)
and, cos(ωt+α)=-cos(ωt)
also we get sin(ωt)=12
sin(ωt)cosα+cos(ωt)sinα=sin(ωt)
or, cosα+cot(ωt)sinα=1
sin(ωt)=1/2 or cosec(ωt)=2 or cosec2(ωt)=4 or cot2(ωt)=3 or cot(ωt)=√3 or tan(ωt)=1/√3
so, cosα+√3sinα=1
using the second set of eqns,
sin(ωt)sinα-cos(ωt)cosα=cos(ωt)
or, tan(ωt)sinα-cosα=1
or, sinα-√3cosα=√3
the 2 bolded eqns on solving give 4sinα=2√3
sinα=√3/2
gives α=60° or 120°
but α can't be 60° by eqn sin(ωt)=√32
thus α=120°