Question

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

Answer 1

The time 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..................eqn1

also for SHM

d²x/dt² + ω²x = 0

=> d²x/dt² = -ω²x

=> force F = ma = m*d²x/dt² = -mω²x............eqn2

from 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 ∝a

Hence the time period is directly proportional to the amplitude a.

Answer 2

Answer:

Explanation:The answer for the above question is

Time period is inversely proportional to square root of amplitude