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Oscillations

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Question : 24 of 25
Marks: +1, -0
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of body when the displacement is (a) 5 cm, (b) 3 cm and (c) 0 cm.
Solution:  
Here,r=5 cm=0.05 mr = 5 \text{ cm} = 0.05 \text{ m} ; T=0.2 sT = 0.2 \text{ s}; ω=2πT=2π0.2=10π rad s−1\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi \text{ rad s}^{-1}
When displacement is y, then acceleration, a=−ω2ya=-\omega^{2} y
velocity, v=ωr2−y2v=\omega \sqrt{r^{2}-y^{2}}
Case (a) When y = 5 cm = 0.05 m
a=−(10π)2×0.05=−5π2 m s−2,a=-(10\pi)^{2}\times 0.05=-5\pi^{2}\,\text{m}\,\text{s}^{-2},
v=10π(0.05)2−(0.05)2=0v=10\pi\sqrt{(0.05)^{2}-(0.05)^{2}}=0
Case (b) When y = 3 cm = 0.03 m
a=−(10π)2×0.03=−3π2 m s−2,a=-(10\pi)^{2}\times 0.03=-3\pi^{2}\,\text{m}\,\text{s}^{-2},
v=10π(0.05)2−(0.03)2v=10\pi\sqrt{(0.05)^{2}-(0.03)^{2}} =10π×0.04=0.4π m s−1=10\pi\times 0.04=0.4\pi\,\text{m}\,\text{s}^{-1}
Case (c) When y = 0
a=−(10π)2×0=0,a=-(10\pi)^{2}\times 0=0,
v=10π(0.05)2−(0)2v=10\pi\sqrt{(0.05)^{2}-(0)^{2}} =10π×0.05=0.5π m s−1=10\pi\times 0.05=0.5\pi\,\text{m}\,\text{s}^{-1}
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