Ncert Exercises & Solutions

14.7 The motion of a particle executing simple harmonic motion is described by the displacement function, x(t)=Acos(ωt+φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π per sec. If instead of the cosine function, we choose the sine function to describe the SHM: x = Bsin(ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Initially, at t = 0 :

Displacement, x = 1 cm

Initial velocity, v = ω cm / \sec .

Angular frequency, ω = π rad / s^{- 1}

It is given that :

x (t) = A \cos (ωt + ϕ)

At t = 0, 1 = A \cos (ω \times 0 + ϕ) = Acosϕ

Acosϕ = 1 (i)

Velocity, v = \frac{dx}{dt}

v = - Aωsin (ωt + ϕ)

At t = 0, 1 = - Asin (ω \times 0 + ϕ) = - Asinϕ

Asinϕ = - 1 (ii)

Squaring and adding equations (i) and (ii), we get :

A^{2} (\sin^{2} ϕ + \cos^{2} ϕ) = 1 + 1

A^{2} = 2

∴ A = \sqrt{2} cm

Dividing equation (ii) by equation (i), we get :

tanϕ = - 1

∴ ϕ = \frac{3 π}{4}, \frac{7 π}{4}, . . . . . . . . . . .

If the SHM is given as :

x = B \sin (ωt + α)

Putting the given values in this equation, we get :

At t = 0, 1 = Bsin [ω \times 0 + α]

Bsinα = 1 (iii)

Velocity, v = ωBcos (ωt + α)

Substituting the given values, we get :

At t = 0, π = πBcosα

Bcosα = 1 (iv)

Squaring and adding equations (iii) and (iv), we get :

B^{2} [\sin^{2} α + \cos^{2} α] = 1 + 1

B^{2} = 2

∴ B = \sqrt{2} cm

Dividing equation (iii) by equation (iv), we get :

\frac{Bsinα}{Bcosα} = \frac{1}{1}

tanα = 1 \Rightarrow tanα = \frac{π}{4}

∴ α = \frac{π}{4}, \frac{5 π}{4}, . . . . . . . . . .

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