A 1.50-kg mass on a spring has displacement as a function of time given by $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{40}\mathbf{}\mathit{c}\mathit{m}\mathbf{)}\mathit{c}\mathit{o}\mathit{s}\left[\left(4.16rad/s\right)t-2.42\right]$.

Find (a) the time for one complete vibration; (b) the force constant of the string; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at t = 1.00 s ; (f) the force on the mass at that time.

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,

$x\left(t\right)=A\cos (\omega t+\varphi )$ …(1)

Here, x(t) 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

The displacement of mass on a spring is given by,

$x\left(t\right)=(7.40cm)\cos [(4.16\mathrm{rad}/\mathrm{s})\mathrm{t}-2.42]$ …(2)

Compare equation. (1) with equation. (2)

role="math" localid="1668089719939" $\begin{array}{ll}A=7.40cm=0.074cm& \\ \omega =4.16rad/s& \\ \varphi =-242& \end{array}$

(a) The relation between time period and angular frequency is given by,

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

Substitute values in above expression to get the time for one complete vibration

$$\begin{gathered} T=\frac{2\mathrm{\pi }}{4.16rad/s} \\ =1.51s \end{gathered}$$

Therefore, the time for one complete vibration is 1.51 s.

(b) 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

$\begin{array}{l}k=(1.50\mathrm{kg}){(4.16rad/s)}^2\\ =25.99N/m\\ \approx 26N/m\end{array}$

Therefore, the force constant of the spring is 26 N/m.

(c) The maximum speed of the mass is given by,

$v_{max}=A\omega$

Substitute the values in above expression to get the maximum speed of the mass

$$\begin{gathered} v_{max}=\left(0.074m\right)(4.16rad/s) \\ =0.31m/s \end{gathered}$$

Therefore, the maximum speed of the mass is 0.31 m/s.

(d) The force is at its maximum value when the mass reaches its maximum displacement.

The force due to spring having spring constant is given by,

F= -kx

At the maximum displacement of the mass, x = A. Therefore, above expression becomes

$\left|F\left|=\right|-kA\right|$

Substitute values in above expression to find the maximum force on the mass

$$\begin{gathered} \left|F\right|=|-(26N/m\left)\right(0.074\mathrm{m}\left)\right| \\ =1.92N \end{gathered}$$

Therefore, the maximum force on the mass is 1.92 N.

(e) The given equation fordisplacement of mass on a spring is

$x\left(t\right)=(7.40cm)\cos [(4.16\mathrm{rad}/\mathrm{s})\mathrm{t}-2.42]$

Substitute t = 1.00 s in above expression and convert the units to their S.I. units $$\begin{gathered} x\left(t\right)=(0.074m)\cos [(4.16\mathrm{rad}/\mathrm{s})\left(1.00s\right)-2.42] \\ =-0.0125m \end{gathered}$$

Therefore, the position of the mass is - 0.0125 m.

The velocity of the mass can be calculated by differentiating the expression of displacement with respect to time.

$v_x=\frac{dx}{dt}$

Substitute$x\left(t\right)=(0.074m)\cos [(4.16\mathrm{rad}/\mathrm{s})t-2.42]$in above expression to get the velocity of mass

$\begin{array}{r}{v}_x=\frac{d\left(\right(0.074\mathrm{m}\left)\cos \right[(4.16\mathrm{rad}/\mathrm{s})t-2.42\left]\right)}{dt}\\ =-0.31\sin (4.16t-2.42)\end{array}$

Substitute t = 1.00 s in above expression

$\begin{array}{r}{v}_x=-0.31\sin \left(4.16\right(1)-2.42)\\ =-0.31\mathrm{m}/\mathrm{s}\end{array}$

Therefore, velocity of the mass is - 0.31 m/s.

The acceleration of mass can be calculated by differentiating the expression of velocity with respect to time.

$a_x=\frac{dx}{dt}$

Substitute${\mathrm{v}}_x=-0.31\sin (4.16t-24)$in above expression to get the acceleration of the mass

$$\begin{gathered} a_x=\frac{d\left(-0.31\sin \left(4.16t-2.42\right)\right)}{dt} \\ =-1.28\cos \left(4.16t-2.42\right) \end{gathered}$$

Substitute t = 1.00 s in above expression

$$\begin{gathered} a_x=-1.28\cos \left(4.16\left(1\right)-2.42\right) \\ =0.22m/s^2 \end{gathered}$$

Therefore, the acceleration of the mass is 0.22m/s².

(f) The force on the mass at that time is given by,

$F=m{a}_x$

Substitute values in above expression to get the force on mass at time t = 1.00 s

$$\begin{gathered} F=(1.50\mathrm{kg})(0.22\mathrm{m}/s^2) \\ =0.33N \end{gathered}$$

Therefore, the force on the mass at that time is 0.33 N.