Question

A 0.10 kgblock oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by$\mathit{x}\mathbf{=}\mathbf{\left(}\mathbf{10}\mathbf{cm}\mathbf{\right)}\mathbf{cos}\mathbf{[}\left(10\mathrm{rad}/s\right)\mathbf{t}\mathbf{+}\mathbf{\tau \tau }\mathbf{/}\mathbf{2}\mathbf{rad}\mathbf{]}$. (a) What is the oscillation frequency? (b) What is the maximum speed acquired by the block? (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what value of x does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?

Short answer

1. The oscillation frequency is 1.6 Hz.
2. The maximum speed acquired by the block is 1.0 m/s.
3. At,0m the maximum speed value occurs.
4. The magnitude of the maximum acceleration is 10m/s.
5. At, 0.1 the maximum acceleration value occurs.
6. The force that is to be applied to the spring is -10xN.

Step-by-step solution

The given data

• The mass of the block, m=0.1kg.
• The angular frequency of the oscillations,$\omega =10\mathrm{rad}/\mathrm{s}$.
• The amplitude of the oscillations,$x_m=10\mathrm{cm}\mathrm{or}0.1\mathrm{m}$.
• The phase constant, $\varphi =\frac{\pi }{2}$.

 Step 2: Understanding the concept of Simple Harmonic Motion

Using the expression of SHM for maximum velocity and maximum acceleration, we can find their respective values and at which position they occur.

Using the expression for angular frequency and force we can find frequency and force.

Formula:

The frequency of the oscillations, $f=\omega /2\tau \tau$ (i)

The maximum velocity of the oscillation,$v_{max}=\omega x_m$ (ii)

The magnitude of the maximum acceleration,$\left|a_{max}\right|={\omega }^2{x}_m$ (iii)

The force of spring as per Hook’s law, $F=kx=m{\omega }^2{x}_m$ (iv)

a) Calculation of oscillation frequency

Using equation (i), the value of oscillation frequency is given as:

$\begin{array}{l}\mathrm{f}=\frac{10\mathrm{rad}/\mathrm{s}}{2\times 3.14}\\ =1.592\mathrm{Hz}\\ \approx 1.6\mathrm{Hz}\end{array}$

Hence the value of frequency is 1.6 Hz.

b) Calculation of maximum speed

Using equation (ii), the maximum speed is given as:

$$\begin{gathered} v_{max}=10rad/s\times 0.1\mathrm{m} \\ =1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of maximum speed is 1.0 m/s.

c) Calculation of position at which maximum speed occurs

$X\left(v_{max}\right)$: The maximum speed occurs at t=0.

$X\left(v_{max}\right)=10\cos \left[\frac{\tau \tau }{2}\right]$

Hence, the required position occurs at 0m.

d) Calculation of maximum acceleration

Using equation (iii), the magnitude of maximum acceleration is given as:

$$\begin{gathered} \left|a_{max}\right|={\left(10\mathrm{rad}/\mathrm{s}\right)}^2\times 0.1\mathrm{m} \\ =10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of maximum acceleration is $10\mathrm{m}/{\mathrm{s}}^2$.

e) Calculation of the position at which the maximum acceleration occurs

$X\left(a_{max}\right)$: The magnitude of acceleration is greatest at extreme positions.

$$\begin{gathered} X\left(a_{max}\right)=x_m \\ =0.1\mathrm{m} \end{gathered}$$

Hence, the required position is 0.1 m.

f) Calculation of the force

We know that the spring constant is given as:

$$\begin{gathered} \mathrm{k}={\mathrm{m\omega }}^2 \\ =10\mathrm{kg}\times 110\mathrm{rads}{)}^2 \\ =10\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, as per equation (iv), we have the force on the block as:

$F=-10\times \mathrm{N}$

Hence, the value of the force is -10xN.