Question

A frictionless block of mass 2.10 kg is attached to an ideal spring with force constant 250 N/m . At t=0 the spring is neither stretched nor compressed and the block is moving in the negative direction at a speed of 13.0 m/s .

Part AFind the amplitude.SubmitMy AnswersGive UpPart BFind the phase angle.

Answer 1

Step-by-step explanation:

Given

mass of block
`m=2.12 kg`

spring constant
`k=250 n/m`

speed of block at t=0 is
`v=13 m/s`

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

`\omega _n=\sqrt{(k)/(m)}`

`\omega _n=\sqrt{(250)/(2.12)}`

`\omega _n=√(117.92)`

`\omega _n=10.85 rad/s`

Suppose
`x=A\cos (\omega _nt+\phi )`

where
`\phi =Phase\ difference`

at t=0 spring is neither compressed

Amplitude is given by

`A=\sqrt{x^2+((v)/(\omega ))^2}`

at t=0 x=0 i.e. at mean Position

`A=√(0+(1.198)^2)`

`A=1.198 m`

at
`t=0`

`x=A\cos (\omega _nt+\phi )`

`x=1.198\cos (0+\phi )`

`0=\cos (\phi )`

therefore
`\phi =(\pi )/(2)`