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A particle of mass m is executing oscillations about the origin on the x axis.Its potential energy is u=k[x] where k is a positive constant.If the amplitude of oscillation is a then its time period t is directly proportional to |
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Answer» e PERIOD is directly PROPORTIONAL to the AMPLITUDE a.Explanation :LET the eqn of the SHM , X = asinωt Given,U = k[x]=> Force F = -dU/dx = -k..................eqn1also for SHMd²x/dt² + ω²x = 0=> d²x/dt² = -ω²x=> force F = ma = m*d²x/dt² = -mω²x............eqn2from eqn1 and eqn2 we get,k = mω²x=> ω = √(k/mx) = √(k/m*asinωt)Time period of oscillation,T = 2π/ω=> T = 2π/√(k/m*asinωt) = 2π x √m*asinωt/k=> T ∝aHence the time period is directly proportional to the amplitude a. |
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