Question

Figure below gives the position of a 20 gblock oscillating in SHM on the end of a spring. The horizontal axis scale is set by${\mathbf{t}}_{\mathbf{s}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{ms}$.

1. What is the maximum kinetic energy of the block?
2. What is the number of times per second that maximum is reached? (Hint: Measuring a slope will probably not be very accurate. Find another approach.)

Short answer

1. The maximum kinetic energy of the block is 1.2 J.
2. The number of times per second that maximum is reached are 50 o scil/sec.

Step-by-step solution

The given data

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

Understanding the concept of oscillations

Calculating angular frequency $\mathit{\omega }$, 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)

Calculation of the maximum kinetic energy of the block

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 $\omega$is given as:

$$\begin{gathered} \omega =\frac{2\tau \tau }{T} \\ =\frac{2\times 3.14}{0.040} \\ =157\mathrm{rad}/\mathrm{s} \end{gathered}$$

But, velocity of the wave is given as:

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

$$\begin{gathered} {\left(KE\right)}_{max}=\frac{1}{2}m{\left(\omega x_x\right)}^2 \\ =\frac{1}{2}m{\omega }^2{x_m}^2 \\ =\frac{1}{2}\times 0.020\times {157}^2\times 0.{070}^2 \\ =1.2\mathrm{J} \end{gathered}$$

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

(b) Calculation of number of times per second where maximum is reached

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$2\times 25=50$

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