Question 254: Oscillations

Question

A vertical spring oscillates with period 6 second with mass $m$ is suspended from it. When the mass is at rest, the spring is stretched through a distance of (Take, acceleration due to gravity, $\mathrm{g}=\pi^2=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Given:
Period of oscillation, $T = 6$ s
Mass suspended, $m$
Acceleration due to gravity, $g = \pi^2 = 10\ 	ext{m/s}^2$
Let the extension in the spring at equilibrium (rest position) be $x$.
Step 1: Relation between period and spring constant
The period for a spring-mass system:
$$ T = 2\pi \sqrt{\frac{m}{k}} $$
where $k$ is the spring constant.
Step 2: Express $k$ in terms of $m$ and $T$
Squaring both sides and rearranging for $k$,
$ T = 2\pi \sqrt{\frac{m}{k}} $
$ \Rightarrow \frac{T}{2\pi} = \sqrt{\frac{m}{k}} $
$ \Rightarrow \left(\frac{T}{2\pi}ight)^2 = \frac{m}{k} $
$ \Rightarrow k = \frac{m}{\left(\frac{T}{2\pi}ight)^2} $
Step 3: Extension at equilibrium
At rest,
$ mg = kx $
$ \Rightarrow x = \frac{mg}{k} $
Step 4: Substitute value of $k$
From above,
$ x = \frac{mg}{k} $
$ = mg 	imes \frac{\left(\frac{T}{2\pi}ight)^2}{m} $
$ = g \left(\frac{T}{2\pi}ight)^2 $
Step 5: Plug the values
Given $T = 6$ s and $g = \pi^2 = 10$,
$ x = 10 \left( \frac{6}{2\pi} ight)^2 $
$ = 10 \left( \frac{3}{\pi} ight)^2 $
$ = 10 	imes \frac{9}{\pi^2} $
But $\pi^2 = 10$, so:
$ x = 10 	imes \frac{9}{10} = 9\ 	ext{m} $
Final Answer:
Option D: 9 m

Answer

10 m

Answer

3 m

Answer

6 m

Answer

9 m

Oscillations

A vertical spring oscillates with period 6 second with mass m is suspended from it. When the mass is at rest, the spring is stretched through a distance of (Take, acceleration due to gravity, \mathrm{g}=\pi^2=10 \mathrm{~m} / \mathrm{s}^2 )

Choose the correct answer: