Answer to Question #97746 in Classical Mechanics for Surej

Question

A 1kg mass hangs at the end of a spring, stretches the spring by a distance X from
the equilibrium position. If the mass is increased by 9gm, the displacement
increases by 10cm.
a) what is the force constant of the spring?
b) If the spring vibrates in SHM, find the ration of frequencies in the two cases.

Answer 1

a)Using the Hooke's law $\left| {\vec F} \right| = k\Delta l$ where $\Delta l$ is the displacement from the equilibrium and $k$ is the force constant of the spring and gravitational force $\left| {\vec F} \right| = mg$ we get

$$mg = k\Delta l$$

Write this equation for the first and the second case we get (we shall use $g \approx 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]$ )

$$\begin{cases}1[{\text{kg}}] \cdot 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}] = k \cdot X[{\text{m}}] \\ (1 + 9 \cdot {10^{ - 3}})[{\text{kg}}] \cdot 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}] = k \cdot (X + {10^{ - 1}})[{\text{m}}]\end{cases}$$

Substract the first from the second we get

$$9 \cdot {10^{ - 3}}[{\text{kg}}] \cdot 10[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}] = k \cdot {10^{ - 1}}[{\text{m}}]$$

Thus

$$k = 0.9[\frac{{{\text{kg}}}}{{{{\text{s}}^{\text{2}}}}}]$$

b)The angular frequency in SHM has the form

$$\omega = \sqrt {\frac{k}{m}}$$

Thus

$$\frac{{{\omega _2}}}{{{\omega _1}}} = \frac{{\sqrt {\frac{k}{{{m_2}}}} }}{{\sqrt {\frac{k}{{{m_1}}}} }} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} = \sqrt {\frac{1}{{1 + 9 \cdot {{10}^{ - 3}}}}} = \sqrt {\frac{{1000}}{{1009}}} \approx 0.9955$$