Question

The period of a particular spring-mass oscillator is $\mathbf{1}$ when the amplitude is$\mathbf{5}\mathbf{}\mathit{c}\mathit{m}$ . (a) what would be the period if we doubled the mass? (b) What would be the period if we replaced the original spring with a spring that is twice as stiff (keeping the original mass)? (c) What would be the period if we cut the original spring in half and use just one of the pieces (keeping the original mass)? (d) What would be the period if we increased the amplitude of the original system to$\mathbf{10}\mathbf{}\mathit{c}\mathit{m}$ , so that the total distance traveled in one period is twice as large? (e) What would be the period if we took the original system to a massive planet where$\mathit{g}\mathbf{=}\mathbf{25}\mathbf{}\mathit{N}\mathbf{/}\mathit{k}\mathit{g}$ ?

Short answer

a) Time period increases with$\sqrt{2}$ time.

b) Time period decreases with$2$ time.

c) Time period will be the same.

d) Time period will be the same.

e) Time period decrease.

Step-by-step solution

Identification of given data

Amplitude, $A=5cm$

Period, $T=1s$

Spring-mass system oscillator

If the masses and springs are the same, both vertical and horizontal spring-mass systems without friction oscillate equally around an equilibrium point. However, when it comes to vertical springs, keep in mind that gravity extends or compresses the spring beyond its natural length to get it to the equilibrium position.

Expression for the time period for the spring-mass system,

$T=\frac{2\mathrm{\pi }}{\omega }...................................\left(1\right)$

Where is period, is the angular frequency.

$\omega =\sqrt{\frac{k}{m}}........................................\left(2\right)$

Where$k$ is the stiffness of spring,$m$ is the mass of the block.

Put the equation number 2 in equation (1), and we get,

$T=2\mathrm{\pi }\sqrt{\frac{m}{k}}.......................................\left(3\right)$

Calculating the time period

Part a)

If we doubled the mass,

$M=2m$

Then from the equation (3),

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

So, the time period increases by the root two times.

Part b)

The time period, if we doubled the stiffness of spring,

$K=2k$then from the equation (3)

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

So, the time period decreases by the root two times.

Part c)

From the above equation, it is independent of the length, so if we cut the original spring in half and use just one of the pieces. It is independent of time. So that time period will be the same.

Part d)

If we increased the amplitude of the original system to 10 cm, as the amplitude is independent of the time period, that time period would be the same.

Part e)