91影视

• 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.

A particle with mass m that moves under the influence of Hooke鈥檚 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,

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

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" $T=2蟿蟿\sqrt{\left(K/m\right)}$ (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)

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

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

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.

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.

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

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

$=0.9\mathrm{s}$

Hence, the value of the period is 0.9 s.

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

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

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