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Academic session 2026-27, Chapter 13 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Physics - Chapter Oscillations. This page offers a step-by-step solution to the specific question from Exercise 1, Question 7:
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]
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 7: The motion of a body in simple harmonic motion is given by the displacement function, x (t) = A c....
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