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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