Question 16: Oscillations

Question

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2 / v_1$ is $\_\_\_\_$ .

Accepted Answer

When a mass $m$ is attached to a spring with spring constant $k$, the restoring force is given by Hooke's law
$$ \mathrm{F}=-\mathrm{kx} $$
According to Newton's Second Law,
$$ \mathrm{m} \frac{\mathrm{~d}^2 \mathrm{x}}{\mathrm{dt}^2}=-\mathrm{kx} \Rightarrow \frac{\mathrm{~d}^2 \mathrm{x}}{\mathrm{dt}^2}+\frac{\mathrm{k}}{\mathrm{~m}} \mathrm{x}=0 $$
Comparing this to the standard SHM equation $\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}+\omega^2 \mathrm{x}=0$, we find the angular frequency as,
$$ \omega=\sqrt{\frac{\mathrm{k}}{\mathrm{~m}}} $$
Since frequency $v=\frac{\omega}{2 \pi}$, the frequency of oscillation is:
$$ v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{~m}}} $$
$$\Rightarrow $$ $$ \mathrm{v} \propto \sqrt{\mathrm{k}} $$
A spring constant k is inversely proportional to the natural length L of the spring $\left(\mathrm{k} \propto \frac{1}{\mathrm{~L}}ight)$.
$$ \mathrm{k} \cdot \mathrm{~L}=	ext { constant } $$
Let the initial length be $\mathrm{L}_1$ and initial spring constant be $\mathrm{k}_1$.
The new length is $L_2=\frac{L_1}{2}$.
Let the new spring constant is $\mathrm{k}_2$.
$$ \mathrm{k}_1 \mathrm{~L}_1=\mathrm{k}_2 \mathrm{~L}_2 \Rightarrow \mathrm{k}_1 \mathrm{~L}_1=\mathrm{k}_2\left(\frac{\mathrm{~L}_1}{2}ight) $$
$$\Rightarrow $$ $\mathrm{k}_2=2 \mathrm{k}_1$
Using the relation between spring constant and frequency of oscillation,
$$ \frac{v_2}{v_1}=\frac{\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}}{\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}} $$
$$\Rightarrow $$ $$ \frac{v_2}{v_1}=\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}} $$
Substituting $\mathrm{k}_2=2 \mathrm{k}_1$ :
$$ \frac{v_2}{v_1}=\sqrt{\frac{2 \mathrm{k}_1}{\mathrm{k}_1}} $$
$$\Rightarrow $$ $$ \frac{v_2}{v_1}=\sqrt{2} $$
Therefore, the value of the ratio $\frac{v_2}{v_1}$ is $\sqrt{2}$.
Hence, the correct option is (C).

Answer

1

Answer

2

Answer

$\sqrt{2}$

Answer

$\sqrt{3}$

Oscillations

The frequency of oscillation of a mass m suspended by a spring is v_1. If the length of the spring is cut to half, the same mass oscillates with frequency v_2. The value of v_2 / v_1 is \_\_\_\_ .

Choose the correct answer: