Danelle Albright

2022-01-05

A 0.500-kg mass on a spring has velocity as a function of time given by ${v}_{x}\left(t\right)=\left(3.60c\frac{m}{s}\right\}\right)\mathrm{sin}\left[\left(4.71ra\frac{d}{s}\right)t-\left(\frac{\pi }{2}\right)\right]$. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

Ben Owens

Step 1
Given that the mass of the object is $m=0.5kg$.
The velocity of the object as a function of time given by
${v}_{x}\left(t\right)=-\left(3.6cm{s}^{-1}\right)\mathrm{sin}\left[\left(4.71rad{s}^{-1}\right)t-\frac{\pi }{2}\right]$ ​(1)
Since we know that the displacement in SHM is given by
$x=A\mathrm{cos}\left[\omega t-\varphi \right]$ (2)
By differentiating above equation with respect to time we get the velocity of the object in SHM, so
${v}_{x}\left(t\right)=-A\omega \mathrm{sin}\left[\omega t-\varphi \right]$ (3)
Now by comparing the equation (1) and (3), we get
$A\omega =3.6cm{s}^{-1}$ (4)
$\omega =4.7rad{s}^{-1}$ (5)
and
$\varphi =\frac{\pi }{2}rad$ (6)
(a) The period of SHM is given by
$P=\frac{2\pi rad}{\omega }$
$=\frac{2\pi rad}{4.71rad{s}^{-1}}$
$=1.33s$ (7)

Matthew Rodriguez

Step 2
$A=\frac{3.6cm{s}^{-1}}{\omega }$
$=\frac{3.6cm{s}^{-1}}{4.71ra{d}^{-1}}$
$=0.76cm$ (8)

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