Question

A block weighing 20 Noscillates at one end of a vertical spring for which k=100 N/m; the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched 0.30 mbeyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?

Short answer

1. Net force acting on the block is 10 N.
2. Amplitude of the resulting SHM is 0.10 m.
3. Time period of the Resulting SHM is 0.9s.
4. Maximum kinetic energy of the block as it oscillates is 0.5 J.

Step-by-step solution

The given data

• Weight of the block, w=20 N .
• The spring constant, k=100 N/m .
• The stretch applied to the spring at x=30m, with V=0.

Understanding the concept of simple harmonic motion

A particle with mass m that moves under the influence of Hooke’s law restoring force, $\mathit{F}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}$,exhibits simple harmonic motion. Here, Fis restoring force, kis force constant and x is the displacement from the mean position.

The period T is the time required for one complete oscillation or cycle. It is related to the frequency by,

role="math" localid="1657274364317" $\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathit{f}}$

It is also related to mass (m) and force constant (k) by the formula,

$\mathit{T}\mathbf{=}\mathbf{2}\mathit{\tau }\mathit{\tau }\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

A particle in simple harmonic motion has, at any time, the maximum kinetic energy,

$\mathit{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{{\mathit{v}}^{\mathbf{2}}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$

We can find the net force, period, and maximum kinetic energy by using respective formulae. Maximum displacement can be found by using the fact that at an extreme position velocity is zero.

Formula:

The force of the spring force,$F=kx$ (i)

The period of oscillations,role="math" localid="1657274559045" $T=2\tau \tau \sqrt{\left(K/m\right)}$ (ii)

At the equilibrium position, the maximum kinetic energy (or maximum potential energy) of the system,

$K{E}_{max}=\frac{1}{2}m{v^2}_{max}$ (iii)

a) Calculation of net force on the block

Using equation (i), the upward force applied on the block can be given as:

$$\begin{gathered} F=100\mathrm{N}/\mathrm{m}\times 0.30\mathrm{m} \\ =30\mathrm{N} \end{gathered}$$

This force is acting upward and weight is acting downwards which is 20 N, so the net force is given as:30 N-20 N=10 N .

Hence, the value of the net force is 10 N.

b) Calculation of amplitude

Using equation (ii) and the downward force value of 20 N, the equilibrium position is given as:

$$\begin{gathered} x=\frac{20\mathrm{N}.\mathrm{m}}{100\mathrm{N}} \\ =0.2\mathrm{m} \end{gathered}$$

As at extreme position velocity is zero, x = 0.30 m is an extreme position.

Hence, the maximum displacement is,

$$\begin{gathered} x_m=0.3\mathrm{m}-0.2\mathrm{m} \\ =0.1\mathrm{m} \end{gathered}$$

Hence, the value of amplitude is 0.1 m.

c) Calculation of period

Using equation (ii) and the given values, we can get the period of the oscillations of the block as:

$$\begin{gathered} T=2\times 3.14\sqrt{\left(\frac{W/g}{k}\right)}\left(\because m=w/g\right) \\ =2\times 3.14\sqrt{\left(\frac{20\mathrm{N}/9.8\mathrm{m}/{\mathrm{s}}^2}{100\mathrm{N}.\mathrm{m}}\right)} \end{gathered}$$

$=0.9\mathrm{s}$

Hence, the value of the period is 0.9 s.

d) Calculation of the maximum kinetic energy

Using equation (iii), we can get the value of maximum kinetic energy at the equilibrium as:

$$\begin{gathered} K{E}_{max}=\frac{1}{2}\times 100\mathrm{kg}{\left(0.10\mathrm{m}/\mathrm{s}\right)}^2 \\ =0.5\mathrm{J} \end{gathered}$$

Hence, the value of maximum kinetic energy is 0.5 J.