Question

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

Short answer

(a)The period of is $1.33s$.

(b) The amplitude is $0.76cm$.

(c) the maximum acceleration of the mass is $16.86cm/s^2$.

(d) The force constant of the spring isrole="math" localid="1664272223666" $11.09N/m$.

Step-by-step solution

Brief about displacement in SHM

Displacement of a body in SHM can be defined as the net distance travelled by the body from its mean or equilibrium position.

Mathematically it is given by,

$x=A\cos \left(\omega t-\varphi \right)..........\left(1\right)$

Here, x is the displacement of body

A is the amplitude of oscillation

$\omega$is the angular frequency

t is the time

$\varphi$is the phase angle

Differentiate equation. (1) with respect to time

$v_x\left(t\right)=-A\sin \omega \left(\omega t-\varphi \right)..........\left(2\right)$

Calculate angular velocity and phase angle

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

$v_x\left(t\right)=-\left(3.60cm/s\right)\sin \left[\left(4.71rad/s\right)t-\left(\pi /2\right)\right]........\left(3\right)$

Compare equation. (2) and (3)

$$\begin{gathered} A\omega =3.60cm/s=0.036m/s \\ \omega =4.71rad/s.......\left(4\right) \\ \varphi =\frac{\pi }{2}rad \end{gathered}$$

Calculate the period of SHM

(a) The period of SHM is given by,

$T=\frac{2\pi }{\omega }$

Substitute values in above expression to get the period of SHM

$$\begin{gathered} T=\frac{2\pi }{4.71rad/s} \\ =1.33s \end{gathered}$$

Therefore, the period of SHM is $1.33s$.

Determine the amplitude of SHM

(b) From equation. (4),

Substitute the value of to get the amplitude

$$\begin{gathered} A=\frac{0.036m/s}{4.71rad/s} \\ =0.76\times {10}^{-2}m \\ =0.76m \end{gathered}$$

Therefore, the amplitude of SHM is 0.76 cm.

Calculate the maximum acceleration of the mass

(c) The acceleration is maximum when displacement is equal to the amplitude of SHM. Therefore, the expression for maximum acceleration of the mass will be

${\mathrm{a}}_{max}={\omega }^{2}A$

Substitute values in above expression to get maximum acceleration

$$\begin{gathered} {\mathrm{a}}_{\max }={\left(4.71\mathrm{rad}/\mathrm{s}\right)}^2\left(0.76\times {10}^{-2}\mathrm{m}\right) \\ =16.86\times {10}^{-2}\mathrm{m}/{\mathrm{s}}^2 \\ =16.86\mathrm{cm}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the maximum acceleration of the mass is $16.86cm/s^2$.

Calculate the force constant of the spring

(d) The angular frequency in SHM is given by,

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

Here, is the spring constant or force constant of the spring.

Rewrite above expression for k.

$k=m{\omega }^2$

Substitute the values in above expression to get the force constant of the spring

$$\begin{gathered} k=\left(0.5kg\right){\left(4.71rad/s\right)}^2 \\ =11.09N/m \end{gathered}$$

Therefore, the force constant of the spring is $11.09N/m$.