Question

A particular spring-mass oscillator oscillates with period T. Write out the general equation for the period of such an oscillator to use as a guide when answering the following questions. (a) If you double the mass but keep the stiffness the same, by what numerical factor does the period change? (That is, if the original period was T and the new period is bT. what is b?) (b) If, instead, you double the spring stiffness but keep the mass the same, what is the factor b? (c) If, instead, you double the mass and also double the spring stiffness, what is the factor b? (d) If, instead, you double the amplitude (keeping the original mass and spring stiffness), what is the factor b?

Short answer

a) If we double the mass but keep the stiffness the same, the period change by the factor $\sqrt{2}$.

b) If we double the spring stiffness but keep the mass the same, the period change by the factor$\frac{1}{\sqrt{2}}$ .

c) If we double the mass and also double the spring stiffness, the factor is 1.

d) If we double the amplitude, the factor is 1.

Step-by-step solution

Spring-mass oscillator

When the spring is allowed to oscillate, this potential energy is released. When the spring returns to its equilibrium, all energy is converted to kinetic energy, and the maximum speed is achieved.

General equation for the period

The general equation for the period of the oscillator is written by,

$$\begin{gathered} T=\frac{2\mathrm{\pi }}{\omega } \\ \omega =\sqrt{\frac{k}{m}} \\ T=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \end{gathered}$$

Where $T$is period, $\omega$is the angular frequency, $k$is the stiffness of spring, and$m$ is the mass of the block.

Calculating the new period factor

Part a)

If we double the mass but keep the stiffness the same, then the period is

$$\begin{gathered} T\text{'}=2\mathrm{\pi }\sqrt{\frac{2\mathrm{m}}{\mathrm{k}}} \\ \mathrm{T}\text{'}=\sqrt{2\times 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}} \\ \mathrm{T}\text{'}=\sqrt{2}\mathrm{T} \end{gathered}$$

Compare to $T\text{'}=\sqrt{2}T$

$b=\sqrt{2}$

Thus, if we double the mass but keep the stiffness the same, the period change by the factor $\sqrt{2}$.

Part b)

If we double the spring stiffness but keep the mass the same, then the period is

$$\begin{gathered} T\text{'}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}} \\ \mathrm{T}\text{'}=\frac{1}{\sqrt{2}}\times 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \mathrm{T}\text{'}=\frac{\mathrm{T}}{\sqrt{2}} \end{gathered}$$

Compare to $T\text{'}=bT$

$b=\frac{1}{\sqrt{2}}$

Thus,if we double the spring stiffness but keep the mass the same, the period change by the factor$\frac{1}{\sqrt{2}}$ .

Part c)

If we double the mass and also double the spring stiffness, then the period is

$$\begin{gathered} T\text{'}=2\mathrm{\pi }\sqrt{\frac{2\mathrm{m}}{2\mathrm{k}}} \\ \mathrm{T}\text{'}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \mathrm{T}\text{'}=\mathrm{T} \end{gathered}$$

Compare to $T\text{'}=bT$

$b=1$

Thus,if we double the mass and also double the spring stiffness, the factor is 1.

Part d)

If we double the amplitude, then the period will be the same, so that the period change by the factor is 1.