91Ó°ÊÓ

• Mass of the block, m = 20 g or 0.020 kg.
• The horizontal axis scale is set by, $t_s=40.0\mathrm{ms}\mathrm{or}0.40\mathrm{s}$.

Calculating angular frequency $\mathit{Ó\neg }$, we can find the maximum kinetic energy of the block. From the graph, we can find the period Tand using the value of Twe can findthe number of oscillations of times per second that maximum is reached.

Formula:

The maximum kinetic energy of the oscillation,${\left(KE\right)}_{max}=\frac{1}{2}m{v}_m^2$ (i)

The frequency of the oscillation, $f=\frac{1}{T}$ (ii)

Fromthegraph, we have amplitude of oscillations

$$\begin{gathered} x_m=7.0\mathrm{cm} \\ =0.070\mathrm{m} \end{gathered}$$

And,

T = 40.0 m s

= 0.040 s

We know that angular frequency $\mathrm{Ó\neg }$is given as:

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï„}\mathrm{Ï„}}{T} \\ =\frac{2×3.14}{0.040} \\ =157\mathrm{rad}/\mathrm{s} \end{gathered}$$

But, velocity of the wave is given as:

Maximum kinetic energy using equation (i) is given by:

$$\begin{gathered} {\left(KE\right)}_{max}=\frac{1}{2}m{\left(\mathrm{Ó\neg }{x}_x\right)}^2 \\ =\frac{1}{2}m{\mathrm{Ó\neg }}^2{x_m}^2 \\ =\frac{1}{2}×0.020×{157}^2×0.{070}^2 \\ =1.2\mathrm{J} \end{gathered}$$

Hence, the value of maximum kinetic energy is 1.2 J.

Using equation (ii), we get the frequency of oscillation as:

$$\begin{gathered} f=\frac{1}{0.040} \\ =25\mathrm{oscillation}/\mathrm{second} \end{gathered}$$

As maximum K.E is reached twice in a cycle, the number of oscillations per second required to reach that energy are$2×25=50$

Hence, the required number of times per second is 50 oscillation / second