A simple harmonic oscillator consists of a 0.50 kgblock attached to a spring. The block slides back and forth along a straight line on a frictionless surface with equilibrium point x=0. At t=0the block is at x=0and moving in the positive x direction. A graph of the magnitude of the net force$\overrightarrow{\mathbf{F}}$on the block as a function of its position is shown in Fig. 15-55. The vertical scale is set by ${\mathbf{F}}_{\mathbf{S}}\mathbf{=}\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. What are (a) the amplitude and (b) the period of the motion, (c) the magnitude of the maximum acceleration, and (d) the maximum kinetic energy?

• The mass mof the block is, m=0.5 Kg .
• At t=0, block is at x=0 .
• The vertical scale is set by ${\mathrm{F}}_{\mathrm{S}}=75.0\mathrm{N}$.

The amplitude of oscillation is the maximum displacement of the pendulum from the equilibrium position.

From the graph, we can find the amplitude of the oscillations of the block, and taking the slope of the graph, we can find the spring constant. Using this value of k, we can find the period of oscillation. Using this value, we can find the maximum speed and maximum acceleration of the block. Finally, using the calculated value of the maximum speed, we can find maximum kinetic energy.

Formula:

The force of the spring, F =-kx (i)

The period of the oscillations, $T=2\pi \sqrt{\frac{m}{k}}$ (ii)

The angular frequency of a wave, $\omega =\frac{2\pi }{T}$ (iii)

The maximum speed of a wave,$v_m=\omega x_m$ (iv)

The maximum acceleration of the particle,$a_m={\omega }^2{x}_m$ (v)

The maximum kinetic energy of the system,$K_m=\frac{1}{2}m{v_m}^2$ (vi)

Where k is force constant, xis displacement from mean position, mis mass,$\omega$ is angular velocity, $v_m$is maximum velocity, $x_m$is maximum displacement or amplitude, $a_m$is maximum acceleration, $k_m$is kinetic energy.

Fromthe givengraph, we can find that the amplitude of the motion,$x_m=0.30\mathrm{m}$

Hence, the value of the amplitude is 0.30m.

Using equation (i), we can see that the spring constant is given as:

$-k=\left(\frac{F}{x}\right)$

But, from the graph, $\left(\frac{F}{x}\right)$and it is negative. So,

$$\begin{gathered} -k=\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{graph} \\ -k=\left(\frac{75.0\mathrm{N}}{-0.30\mathrm{m}}\right) \\ -\mathrm{k}=(-250) \\ \mathrm{k}=250\mathrm{N}/\mathrm{m} \end{gathered}$$

Now, putting the value of spring constant in equation (ii), the period is given by:

$$\begin{gathered} T=2\times 3.14\sqrt{\frac{0.50\mathrm{kg}}{250\mathrm{N}/\mathrm{m}}} \\ =0.281\mathrm{s} \end{gathered}$$

Hence, the value of period of the oscillations is 0.281 s.

Using the value of period in equation (iii), we get the angular frequency as:

$$\begin{gathered} \omega =\frac{2\times 3.14}{0.281} \\ =22.3488\mathrm{rad}/\mathrm{s} \end{gathered}$$

Now, the maximum speed of the oscillations using equation (iv) is given by:

$$\begin{gathered} v_m=22.3488\times 0.30 \\ =6.705\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the magnitude of the maximum acceleration using equation (v) is given by:

$$\begin{gathered} a_m=22.3488\times 0.30 \\ =149.84\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of maximum acceleration is 149.84 $\mathrm{m}/{\mathrm{s}}^2$.

The maximum kinetic energy of the oscillations using equation (vi) is given by:

$$\begin{gathered} K_m=\frac{1}{2}\times 0.50\times 6.{705}^2 \\ =11.24\mathrm{J} \end{gathered}$$

Hence, the value of maximum kinetic energy is 11.24 J.