Question

A block attached to a spring with unknown spring constant oscillates with a period of $2.0s$.What is the period if

a. The mass is doubled?

b. The mass is halved?

c. The amplitude is doubled?

d. The spring constant is doubled? Parts a to d are independent questions, each referring to the initial situation.

Short answer

(a) When the mass is doubled, the period is $2.83s$

(b) When the mass is halved, the period is $1.41s$

(c) When the amplitude is doubled, it does not affect the time period

(d) When the spring constant is doubled, the period is decreases $t=1.41s$

Step-by-step solution

Find the time when the mass is doubled (part a)

Use angular frequency expression for spring mass system.

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

The periodic time is,

$$\begin{gathered} T=2\pi \sqrt{\frac{m}{k}} \\ \Rightarrow T\propto \sqrt{m} \end{gathered}$$

It Rewritten as

$\frac{T_2}{T_1}=\frac{\sqrt{m_2}}{\sqrt{m_1}}$

$m$is doubled, $m_2=2m_1$

$$\begin{gathered} T_2=T_1\frac{\sqrt{2m_1}}{\sqrt{m_1}} \\ =2\sqrt{2}\mathrm{s} \end{gathered}$$

As a result, the periodic time is raised by $1.4$times.

Find the time when the mass is halved (part b)

$m$is halved then $m_1=2m_2$then,

$$\begin{gathered} T_2=2\times \frac{\sqrt{m_2}}{\sqrt{2m_2}} \\ =\frac{2}{\sqrt{2}} \\ =1.41\mathrm{s} \end{gathered}$$

so the time is$0.707$of initial time.

Find the time when the amplitude is doubled (part c)

Because amplitude is independent of time, changes in amplitude have no effect on time.

Find the time when the spring constant is doubled (part d)

The time constant is inversely proportional to the periodic time,

$\frac{T_2}{T_1}=\frac{\sqrt{k_1}}{\sqrt{k_2}}$

so, $k_2=2k_1$

$$\begin{gathered} T_2=2\times \frac{\sqrt{k_1}}{\sqrt{2k_1}} \\ =\frac{2}{\sqrt{2}} \\ =1.41\mathrm{s} \end{gathered}$$

so the time decreased $0.707$of initial time