Question

Figure below gives the position x(t)of a block oscillating in SHM on the end of a spring$\mathbf{(}{\mathbf{t}}_{\mathbf{s}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ms}\mathbf{)}$.

1. What is the speed of a particle in the corresponding uniform circular motion?
2. What is the magnitude of the radial acceleration of that particle?

Short answer

1. The speed of a particle in the corresponding uniform circular motion is, 11 m/s.
2. The magnitude of the radial acceleration of a particle in the corresponding uniform circular motion is, $1.7\times {10}^3\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

The given data

• The horizontal axis scale is set by,$t_{\mathrm{s}}=40.0\mathrm{ms}\mathrm{or}0.040\mathrm{s}$.
• The position, ${\mathrm{x}}_{\mathrm{m}}=7.0\mathrm{cm}\mathrm{or}0.070\mathrm{m}$.

Understanding the concept of waves of oscillations

First, we find the angular frequency using a given period. Calculating angular frequency, we can find the speed of a particle in the corresponding uniform circular motion and the magnitude of the radial acceleration of a particle in the corresponding uniform circular motion directly using corresponding formulas for SHM.

Formula:

The angular frequency of a wave,$\omega =\frac{2\tau \tau }{\mathrm{T}}$ (i)

Where T is the time period.

The maximum speed of SHM, $v_m=\omega x_m$ (ii)

Where is maximum velocity and $x_m$is maximum displacement.

The maximum acceleration of SHM, $a_m={\omega }^2{x}_m$ (iii)

(a) Calculation of speed of a particle

From the graph, we find that the amplitude of the motion

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

And

T = 40 ms

= 0.040 s

Using equation (i), the angular frequency of the oscillations is calculated by substituting the given values.

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

The maximum speed of the particle is calculated by substituting the value of maximum displacement in the equation (ii) as,

$$\begin{gathered} v_m=157\times 0.070m \\ =11\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of a particle in the corresponding uniform circular motion is 11 m / s.

(b) Calculation of magnitude of the radial acceleration of a particle

The maximum acceleration of the block is calculated by substituting the value of maximum displacement in equation (iii)

$$\begin{gathered} a_m={157}^2\times 0.070\mathrm{m} \\ =1.7\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of the radial acceleration of a particle in the corresponding uniform circular motion is $1.7\times {10}^3\mathrm{m}/{\mathrm{s}}^2$.