Question

Periodic MotionA block of mass M is attached to a horizontal spring with force constant k. It is moving with simple harmonic motion of amplitude A.a) Calculate how much of the energy of the motion is kinetic at x= ¼ A.b) If one adds a mass smoothly in a vertical drop at x=A, calculate what happens to A, T, and w.

Answer 1

Step-by-step explanation:

Given

mass of block is M

Force constant is k

Natural Frequency of oscillation
`\omega _n=\sqrt{(k)/(m)}`

Potential Energy at any instant
`U=(1)/(2)kx^2`

where x is compression in spring

`U=(1)/(2)k((A)/(4))^2`

`U=(1)/(2)k((A)/(16))`

Total Energy
`=(1)/(2)kA^2` when mass is at maximum Position

Total Energy=Kinetic Energy+Potential Energy

kinetic Energy
`=(1)/(2)kA^2-(1)/(32)kA^2`

Kinetic Energy
`=(15)/(32)kA^2`

(b)If one adds a mass smoothly in a vertical drop at
`x=A`

then its amplitude is going to decrease because more mass is added to the system

its natural frequency changed to
`\omega _n' =\sqrt{(k)/(2M)}`

so time Period increases because
`T\cdot \omega =2\pi`

`T=(2\pi )/(\omega _n')`