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प्रश्न
Potential energy of a particle performing linear S.H.M. is 0.1 π2x2 joule. If the mass of the particle is 20 g, find the frequency of S.H.M.
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उत्तर
Data: PE = 0.1 π2x2 J, m = 20 g = 2 × 10−2 kg
PE = `1/2`mω2x2 = `1/2`m (4π2f2)x2
∴ `1/2`m (4π2f2)x2 = 0.1π2x2
∴ 2mf2 = 0.1
∴ f2 = `1/(20(2xx10^-2))` = 2.5
∴ The frequency of SHM is
f = `sqrt(2.5)` = 1.581 Hz
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