1. Tardigrade
  2. Question
  3. Physics
  4. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1 (t) = A sin ω t and x2(t) = Asin( ω t + ( 2 π / 3) ) Adding a third sinusoidal displacement x3(t)=B sin(ω l + Φ) brings the mass to a complete rest. The values of B and Φ are

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Q. A point mass is subjected to two simultaneous sinusoidal
displacements in x-direction, $ x_1 (t) = A sin \omega t $ and
$x_2(t) = Asin\bigg( \omega t + \frac{ 2 \pi }{ 3} \bigg) $ Adding a third sinusoidal
displacement $x_3(t)=B sin(\omega l + \Phi) $ brings the mass to a
complete rest. The values of B and $\Phi $ are

IIT JEEIIT JEE 2011Oscillations

Solution:

Resultant amplitude of $ x_1 \, \, and \, \, x_ 2 $ is A at angle $ \bigg(\frac{\pi}{3}\bigg) \, \, from \, \, A_1$.
To make resultant of $ x_1 ,x_2 \, \, and \, \, x_3 $ to be zero.$ A_3 $ should be
equal to A at angle $ \Phi =\frac{4 \pi }{3} $ as shown in figure.
$\therefore $ Correct answer is (b).
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $
Alternate Solution
It we substitute, $x_1 +x_2 + x_3 =0 $
or $ Asin \omega t + A sin \bigg( \omega t + \frac{ 2 \pi}{3} \bigg) + B sin (\omega t + \Phi ) =0 $
Then by applying simple mathematics we can prove that
$ B= A \, \, and \, \, \Phi = \frac{ 4 \pi}{3}$

Solution Image