Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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A Particle is Subjected to Two Simple Harmonic Motions Given by X1 = 2.0 Sin (100π T) and X2 = 2.0 Sin (120 π T + π/3), Where X is in Centimeter and T in Second. - Physics

Sum

A particle is subjected to two simple harmonic motions given by x1 = 2.0 sin (100π t) and x2 = 2.0 sin (120 π t + π/3), where x is in centimeter and t in second. Find the displacement of the particle at (a) = 0.0125, (b) t = 0.025.

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Solution

Given are the equations of motion of a particle:
 x1 = 2.0sin100 \[\pi\]t

\[x_2  = 2 . 0\sin\left( 120\pi t + \frac{\pi}{3} \right)\]

The Resultant displacement \[\left( x \right)\] will be

x = x1 + x2

\[= 2\left[ \sin\left( 100\pi t \right) + \sin\left( 120\pi t + \frac{\pi}{3} \right) \right]\]

(a) At t = 0.0125 s

\[x = 2\left[ \sin\left( 100\pi  \times 0 . 0125 \right) + \sin\left( 120\pi  \times 0 . 0125 + \frac{\pi}{3} \right) \right]\] 

\[   = 2\left[ \sin  \left( \frac{5\pi}{4} \right) + \sin  \left( \frac{3\pi}{2} + \frac{\pi}{3} \right) \right]\] 

\[   = 2\left[ \left( - 0 . 707 \right) + \left( - 0 . 5 \right) \right]\] 

\[   = 2 \times \left( - 1 . 207 \right) =  - 2 . 41  cm\]

(b) At t = 0.025 s

\[x = 2\left[ \sin\left( 100  \pi \times 0 . 025 \right) + \sin\left( 120\pi \times 0 . 025 + \frac{\pi}{3} \right) \right]\] \[   = 2\left[ \sin\left( \frac{10\pi}{4} \right) + \sin\left( 3\pi + \frac{\pi}{3} \right) \right]\] 

\[   = 2\left[ 1 + \left( - 0 . 866 \right) \right]\] 

\[   = 2 \times \left( 0 . 134 \right) = 0 . 27  \text { cm }\]

 

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 57 | Page 256
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