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?

• 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, $\varphi =\frac{\pi }{2}$.

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)

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.

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.

$X\left(v_{max}\right)$: 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.

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 $10\mathrm{m}/{\mathrm{s}}^2$.

$X\left(a_{max}\right)$: 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.

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.