Question

A 2.0 kgblock executes SHM while attached to a horizontal spring of spring constant 200 N/m.The maximum speed of the block as it slides on a horizontal frictionless surface is 3.0 m/s. What are (a) the amplitude of the block’s motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete 7.0cycles of its motion?

Short answer

1. The amplitude of the block’s motion is 0.3 m .
2. The magnitude of its maximum acceleration 30 m/s².
3. The magnitude of the minimum acceleration is 0.
4. Time to complete 7.0 cycles of its motion is 4.4 s .

Step-by-step solution

The given data

• Mass of the block,$m=2.0\mathrm{kg}$.
• Spring constant,$k=200\mathrm{N}/\mathrm{m}$.
• Maximum speed of the block, $v_{max}=3.0\mathrm{m}/\mathrm{s}$.

Understanding the concept of SHM

We can first find angular frequency from mass and spring constant. We can use the formula of maximum velocity to find amplitude. To find maximum acceleration we used the relation between amplitude and angular frequency. Using the period, we can find the time to complete 7 cycles.

Formulae:

The maximum speed of a body in SHM, $v_{max}=A\omega$ (i)

The maximum acceleration of a body in SHM, $a_{max}=A{\omega }^2$ (ii)

The period of oscillation, $T=\left(\frac{2\pi }{\omega }\right)$ (iii)

The angular frequency of a body in SHM, $\omega =\sqrt{\frac{K}{m}}$ (iv)

(a) Calculation of amplitude

Using equation (iv), we get the angular frequency of the block as:

$$\begin{gathered} \omega =\sqrt{\frac{200\mathrm{N}/\mathrm{m}}{2\mathrm{kg}}} \\ =10\mathrm{rad}/\mathrm{s} \end{gathered}$$

Using equation (i) and the given values, we can get the amplitude of the motion as:

$$\begin{gathered} 3=\mathrm{A}\times 10 \\ \mathrm{A}=0.3\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is $0.3\mathrm{m}$.

(b) Calculation of maximum acceleration

Using equation (ii) and the derived values in part (a), we can get the maximum acceleration of the block as:

$$\begin{gathered} a_{max}=0.3\times {10}^2 \\ =30\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of maximum acceleration is $30\mathrm{m}/{\mathrm{s}}^2$.

(c) Calculation of minimum acceleration

Here is x is the displacement from the mean position. Value of acceleration depends on the value of x. At mean position, value of x is zero. Hence, at mean position, acceleration would be zero. Therefore, using equation (ii), we get the magnitude of minimum acceleration as:

$a_{minm}=0$

Hence, the minimum acceleration of the block is zero.

(d) Calculation of time required to complete 7 cycles of motion

Using equation (iii), the period of oscillations of the block is given as:

$$\begin{gathered} T=\frac{2\pi }{10} \\ =0.2\pi s \end{gathered}$$

Time required for completing 7.0 cycles would be given as:

$$\begin{gathered} T_7=0.2\pi \times 7 \\ =1.4\pi \\ =4.4\mathrm{s} \end{gathered}$$

Hence, the value of required time is $4.4\mathrm{s}$.