Question

A particle of mass 0.50 kg executes a simple harmonic motion under a force `F=-(50Nm^-1)x`. If it crosses the centre of oscillation with a speed of `10ms^-1`, find the amplitude of the motion.

Answer 1

The kinetic energy of the pasrticle when it is at the centre of oscillation si `E=1/2mv^2`
`=1/2(0.50kg)(10ms^-1)^2`
`=25J`
The potential energy is zero here. At the maximum displacement `x=A`, the speed is zero and hence the kinetic energy is zero. The potential energyi here is `1/2kA^`. As there is no loss of energy.
`=1/2kA^2=25J` .............i
The force on the particle is given by
`F=(50Nm^-1)mx`.
Thus the spring constant is `k=50Nm^-1` Equation i gives
`=1/2(50Nm^-1)A^2=25J`
o `A=1m`