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प्रश्न
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) t = 0.0125, (b) t = 0.025.
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उत्तर
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
\[x = 2\left[ \sin\left( 100 \pi \times 0 . 025 \right) + \sin\left( 120\pi \times 0 . 025 + \frac{\pi}{3} \right) \right]\] \[ = 2\left[ \sin\left( \frac{10\pi}{4} \right) + \sin\left( 3\pi + \frac{\pi}{3} \right) \right]\]
\[ = 2\left[ 1 + \left( - 0 . 866 \right) \right]\]
\[ = 2 \times \left( 0 . 134 \right) = 0 . 27 \text { cm }\]
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