Danelle Albright

2022-01-05

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

Ben Owens

Beginner2022-01-06Added 27 answers

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

${v}_{x}\left(t\right)=-\left(3.6cm{s}^{-1}\right)\mathrm{sin}[\left(4.71rad{s}^{-1}\right)t-\frac{\pi}{2}]$ (1)

Since we know that the displacement in SHM is given by

$x=A\mathrm{cos}[\omega t-\varphi ]$ (2)

By differentiating above equation with respect to time we get the velocity of the object in SHM, so

${v}_{x}\left(t\right)=-A\omega \mathrm{sin}[\omega t-\varphi ]$ (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)

Given that the mass of the object is

The velocity of the object as a function of time given by

Since we know that the displacement in SHM is given by

By differentiating above equation with respect to time we get the velocity of the object in SHM, so

Now by comparing the equation (1) and (3), we get

and

(a) The period of SHM is given by

Matthew Rodriguez

Beginner2022-01-07Added 32 answers

Step 2

$A=\frac{3.6cm{s}^{-1}}{\omega}$

$=\frac{3.6cm{s}^{-1}}{4.71ra{d}^{-1}}$

$=0.76cm$ (8)

22+64

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