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A body executing linear S.H.M has a velocity of 3cm/s. When its displacement is 4cm. And a velocity of 4 cm/s. When it's displacement is 3 a) find amplitude and period of oscillation b) if the mass of the body is 100 g. Calculate total energy of oscillation. |
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Answer» uddy,◆ Answer-a = 5 cmT = 6.284 SE = 1.25×10^-4 J◆ Explaination-# Given-v1 = 3 cm/s at x1 = 4 cmv2 = 4 cm/s at x2 = 3 cmm = 100 g = 0.1 KG# Solution-Velocity of the particle in SHM is given by-v = ω √(a^2-x^2)v^2 = ω^2 (a^2-x^2)Therefore,v1^2 = ω^2 (a^2-x1^2)3^2 = ω^2 (a^2-4^2)9 = ω^2 (a^2-16) ...(1)v2^2 = ω^2 (a^2-x2^2)4^2 = ω^2 (a^2-3^2)16 = ω^2 (a^2-9) ...(2)Solving (1) & (2), you'll get a = 5 cmω = 1 rad/sPeriod of oscillation is CALCULATED by-T = 2π/ωT = 2×3.142/1T = 6.284 sTotal energy in oscillation is calculated by-E = 1/2 m(ωa)^2E = 1/2 × 0.1 × (1×0.05)^2E = 1.25×10^-4 JHope this helps... |
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