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Simple Harmonic Motion

Section: Physics

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Question : 11 of 23

Marks: +1, -0

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction,

x_1

(t) = Asinω t and

x_2

\left(\omega t + \frac{2\pi}{3}\right)

. Adding a third sinusoidal displacement

x_3

(t) = B sin (ωt + ϕ) brings the mass to a complete rest. The value of B and ϕ are

[JEE Adv 2011 P2]

$\sqrt{2} A, \frac{3\pi}{4}$
$A, \frac{4\pi}{3}$
$\sqrt{3} A , \frac{5\pi}{6}$
$A, \frac{\pi}{3}$

Solution:

The total displacement is

x =

x_1(t) + x_2(t) + x_3(t)

= 0

A sin ωt +

A\sin\left(\omega t + \frac{2\pi}{3}\right)

+ B sin (ωt + ϕ) = 0

A\sin\omega t\left(1+\cos\frac{2\pi}{3}\right)

+ A cos ωt sin

\frac{2\pi}{3}

+ B sin (ωt + ϕ) = 0

\frac{A}{2} \sin\omega t + \frac{\sqrt{3}}{2} A \cos\omega t

= - B sin ωt cos ϕ - B cos ωt sin ϕ

Therefore,

\frac{A}{2}

= - B cos ϕ (1)

and

\frac{\sqrt{3}}{2}

A = - B sin ϕ (2)

Substituting A from Eq. (1), we get

\sqrt{3} \cos \phi

= - sin ϕ ⇒ ϕ =

\frac{4\pi}{3}

and by back substituting, we get B = A.

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