Class 11 Physics - Chapter Oscillations NCERT Solutions | The motion of a body in simple harmonic
The motion of a body in simple harmonic motion is given by the displacement function,
x (t) = A cos (ωt + φ).
Given that at t = 0, the initial velocity of the body is ω cm/s and its initial position is 1 cm, calculate its initial phase angle and amplitude?
If in place of the cosine function, a sine function is used to represent the simple harmonic motion:
x = B sin (ωt + α),
calculate the body’s amplitude and initial phase considering the initial conditions given above. [Angular frequency of the particle is π/ s]
Given,
Initially, at t = 0:
Displacement, x = 1 cm
Initial velocity, v = ω cm/sec.
Angular frequency, ω = π rad/s
It is given that:
x(t) = A cos( ωt + Φ) . . . . . . . . . . . . . . . . ( i )
1 = A cos( ω x 0 + Φ) = Acos Φ
A cos Φ = 1 . . . . . . . . . . . . . . . . ( ii )
Velocity, v = dx / dt
differentiating equation ( i ) w.r.t ‘t’
v = – Aωsin ( ωt + Φ)
Now at t = 0; v = ω and
=> ω = – Aωsin ( ωt + Φ)
1 = – A sin( ω x 0 + Φ) = -Asin(Φ)
Asin(Φ) = – 1 . . . . . . . . . . . . . . . . . . . . . . . ( iii )
Adding and squaring equations ( ii ) and ( iii ), we get:
A2(sin2 Φ + cos2 Φ) = 1 +1
thus, A =√2
Dividing equation ( iii ) by ( ii ), we get :
tan Φ = -1
Thus, Φ =3π/4 , 7π/4
Now if simple harmonic motion is given as :
x = B sin( ωt + α)
Putting the given values in the equation , we get :
1 = B sin ( ω x 0 + α)
Bsin α = 1 . . . . . . . . . . . . . . . . . . . . ( iv )
Also, velocity ( v ) = ω Bcos (ωt + α)
Substituting the values we get :
π = π B sin α
B sin α = 1 . . . . . . . . . . . . . . . . . . . . ( v )
Adding and squaring equations ( iv ) and ( v ), we get:
B2[ sin2 α + cos2 α] =2
Therefore, B = √ 2
Dividing equation ( iv ) by equation ( v ), we get :
B sin α / B cos α = 1
tan α =1 = tan (π/4)
Therefore, α = π/4 , 5π/4, ......
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