Answer to Question #97219 in Classical Mechanics for Surej

Question

A 40gm mass hangs at the end of a spring. When a 20gm mass is added the spring stretches by an additional 5cm.
a) What is the force constant of the spring?
b) What is the frequency with which the 40gm mass vibrates?

Answer 1

We can find the displacement relative to the equilibrium position using the Hooke's law

$$F = k\Delta l$$

Where $F$ is the magnitude of the force acting on the spring (exclude the spring itself), $\Delta l$ - displacement and $k$ - the force constant of the spring.

a) Let's use that $F = mg$ and write down the Hooke's law equation for the two cases (we let $g \approx 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]$)

$$\begin{cases} 40 \cdot {10^{ - 3}}[{\text{kg}}] \cdot 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}] = k\Delta l \\ (40 + 20)\cdot {10^{ - 3}}[{\text{kg}}] \cdot 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}] = k(\Delta l + 5 \cdot {10^{ - 2}}[{\text{m}}]) \end{cases}$$

From the first equation we've got

$$\Delta l = \frac{{0.4}}{k}[\frac{{{\text{kg}} \cdot {\text{m}}}}{{{{\text{s}}^{\text{2}}}}}]$$

Substitute in to the second

$$0.6[\frac{{{\text{kg}} \cdot {\text{m}}}}{{{{\text{s}}^{\text{2}}}}}] = 0.4[\frac{{{\text{kg}} \cdot {\text{m}}}}{{{{\text{s}}^{\text{2}}}}}] + 5 \cdot {10^{ - 2}}[{\text{m}}] \cdot k$$

Thus

$$k = \frac{{0.2[\frac{{{\text{kg}} \cdot {\text{m}}}}{{{{\text{s}}^{\text{2}}}}}]}}{{5 \cdot {{10}^{ - 2}}[{\text{m}}]}} = 4[\frac{{{\text{kg}}}}{{{{\text{s}}^{\text{2}}}}}]$$

b)The Hooke's law can be written as

$${F_x} = - kx$$

(now we use projection of the force ${F_x}$ to the x-axis, thus we have that minus sign, means that the force is "trying" to return the spring in the equilibrium state, $x$ - displacement)

Then using the Newton's second law

$$m\ddot x = - kx$$

And can be rewritten as

$$\ddot x + \frac{k}{m}x = 0$$

It's general solution is

$$x(t) = A\sin (t\sqrt {\frac{k}{m}} + {\varphi _0})$$

Thus the circular frequency is

$$\omega = \sqrt {\frac{k}{m}} = \sqrt {\frac{{4[\frac{{{\text{kg}}}}{{{{\text{s}}^{\text{2}}}}}]}}{{0.04[{\text{kg}}]}}} = 10[\frac{{\text{1}}}{{{{\text{s}}^{ - 1}}}}]$$

Sometimes under frequency they mean the quantity

$$\nu = \frac{\omega }{{2\pi }} = \frac{5}{\pi }[{\text{Hz}}] \approx 1.6[{\text{Hz}}]$$