refer to the attachment
Answers
answer is in the attachment given above
Answer:
Intially, at t = 0;
Displacement, x=1cmx=1cm
Intial velocity, v=\omega\, cm/sec.v=ωcm/sec.
Angular frequency, \omega = \pi\, rad/s^{-1}ω=πrad/s
−1
It is given that,
x(t) = A\,cos(\omega t+\phi)x(t)=Acos(ωt+ϕ)
1=A\, cos(\omega\times 0+\phi) = A\,cos\, \phi1=Acos(ω×0+ϕ)=Acosϕ
A\, cos\, \phi = 1Acosϕ=1 ...(i)
Velocity, v=dx/dtv=dx/dt
\omega = -A\,\omega\, sin(\omega t+\phi)ω=−Aωsin(ωt+ϕ)
1=-A\, sin(\omega\times 0+\phi)=-A\,sin\,\phi1=−Asin(ω×0+ϕ)=−Asinϕ
A\,sin\,\phi = -1Asinϕ=−1 ...(ii)
Squaring and adding equations (i) and (ii), we get:
A^2(sin^2\phi+cos^2\phi)=1+1A
2
(sin
2
ϕ+cos
2
ϕ)=1+1
A^2=2A
2
=2
\therefore A = \sqrt{2}\,cm∴A=
2
cm
Dividing equation (ii) by equation (i), we get:
tan\,\phi = -1tanϕ=−1
\therefore \phi = 3\pi/4, 7\pi/4∴ϕ=3π/4, 7π/4
SHM is given as:
x= B\, sin(\omega t+\alpha)x=Bsin(ωt+α)
Putting the given values in this equation, we get:
1=B\, sin(\omega\times 0+\alpha]=1+11=Bsin(ω×0+α]=1+1
B\,sin\,\alpha=1Bsinα=1 ...(iii)
Velocity, v=\omega\, cos\,(\omega t+\alpha)v=ωcos(ωt+α)
Substituting the given values, we get:
\pi = \pi \,B\,cos\,\alphaπ=πBcosα
B\,cos\,\alpha = 1Bcosα=1 ...(iv)
Squaring and adding equations(iii) and (iv), we get:
B^2[sin^2\alpha+cos^2\alpha]=1+1B
2
[sin
2
α+cos
2
α]=1+1
B^2=2B
2
=2
\therefore B = \sqrt{2}∴B=
2
Dividing equation (iii) by equation (iv), we get:
B\,sin\,\alpha/B\, cos\, \alpha = 1/1Bsinα/Bcosα=1/1
tan\, \alpha = 1=tan\, \pi/4tanα=1=tanπ/4