A simple harmonic oscillator consists of a block attached to a spring with k=200 N/m. The block slides on a frictionless surface, with an equilibrium point x=0and amplitude 0.20 m. A graph of the block’s velocity v as a function of time t is shown in Fig. 15-60. The horizontal scale is set by${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{s}$. What are (a) the period of the SHM, (b) the block’s mass, (c) its displacement at$\mathbf{t}\mathbf{=}\mathbf{0}$, (d) its acceleration at$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{s}$, and (e) its maximum kinetic energy.

• The spring constant is,$k=200\mathrm{N}/\mathrm{m}$.
• Amplitude of the motion of the block is,$x_m=0.20\mathrm{m}$.
• A graph of block’s velocity vs. time.
• The horizontal scale is set at:$t_s=0.20\mathrm{s}$.

We can find the period of oscillation of the block from the graph. Inserting it into the formula for the period of SHM we can find the mass of the block. We can find the displacement of the graph at t=0 from amplitude. Then we can find its acceleration from maximum amplitude. Lastly, we can find the maximum K.E using the formula for it from amplitude and wavenumber.

Formulae:

The elastic P.E energy of the system,$\frac{1}{2}k{x}^2$ (i)

Maximum acceleration of SHM,$a={\omega }^2{x}_m$(ii)

The maximum K.E of SHM,$K.E.=\frac{1}{2}k{x^2}_m$(iii)

The period of oscillation,$T=2\pi \sqrt{\frac{m}{k}}$ (iv)

We can interpret from the graph that the period of the motion of the block is T=0.20s.

Using equation (iv), the mass of the block can be given as:

$$\begin{gathered} m=\frac{T^{2}k}{4{\pi }^2} \\ =\frac{{(0.20)}^2\left(200\right)}{4{(3.142)}^2} \\ =0.20\mathrm{kg} \end{gathered}$$

Therefore, mass of the block is0.20 kg.

We can interpret from the graph that the velocity of the block is zero at $\mathrm{t}=0\mathrm{s}$.It implies that$x=\underline{+}{x}_m$_.

Since the acceleration is positive, then$ma=-kx$implies value of x must be negative.

So,$x=-x_m$

Therefore, the displacement of block at $t=0$isrole="math" localid="1657275129307" $-0.20\mathrm{m}$.

We can infer from the graph that v=0at $t=0.10\mathrm{s}$. So, acceleration at$t=0.10\mathrm{s}$is,$a=\underline{+}{a}_m$_.From the graph we can interpret that a is negative at$t=0.10\mathrm{s}$. So, using equation (ii), we get the value of acceleration as:

$$\begin{gathered} a=-\frac{k}{m}{x}_m(\because {\omega }^2=\frac{k}{m}) \\ =-\frac{200}{0.20}(0.20) \\ =-200\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of block at t=0.10 s is$-200\mathrm{m}/{\mathrm{s}}^2$.

We note from the graph that, maximum speed is$v_m=6.28\mathrm{m}/\mathrm{s}$

So, maximum K.E of the block using equation (i) is given as:

${\mathrm{KE}}_{\max }=\frac{1}{2}(200\mathrm{N}/\mathrm{m}){(0.20\mathrm{m})}^2=4.0\mathrm{J}$