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

Answer:

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 $\omega _n=\sqrt{\frac{k}{m}}$

$\omega _n=\sqrt{\frac{k}{m}}$

$\omega _n=\sqrt{\frac{250}{2.12}}$

$\omega _n=\sqrt{117.92}$

$\omega _n=10.85 rad/s$

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

where $\phi =Phase\ difference$

at t=0 spring is neither compressed

Amplitude is given by

$A=\sqrt{x^2+(\frac{v}{\omega })^2}$

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

$A=\sqrt{0+(1.198)^2}$

$A=1.198 m$

at $t=0$

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

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

$0=\cos (\phi )$

therefore $\phi =\frac{\pi }{2}$