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A particle performs SHM of amplitude 10 cm. Its maximum velocity during oscillations is 100 cm/s. What is its displacement, when the velocity is 60 cm/s? |
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Answer» Data : A = 10 cm, vmax = 100 cm/s, v = 60 cm/s vmax = ωA = 100 cm/s \(\therefore\) ω = \(\frac{v_{max}}A\) = \(\frac{100}{10}\) = 10 rad/s Let x be the displacement when the velocity is v = 60 cm/s. Then, v = ω \(\sqrt{A^2-x^2}\) \(\therefore\) 60 = 10 \(\sqrt{100-x^2}\) \(\therefore\) 6 = \(\sqrt{A^2-x^2}\) \(\therefore\) 36 = 100 - x2 \(\therefore\) x2 = 64 \(\therefore\) x = ± 8 cm |
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