Question 48: Oscillations

Question

A spring of force constant $15 \mathrm{~N} / \mathrm{m}$ is cut into two pieces. If the ratio of their length is $1: 3$, then the force constant of smaller piece is $\_\_\_\_$ $\mathrm{N} / \mathrm{m}$.

Accepted Answer

The spring constant k of a spring is inversely proportional to the length L of the spring.
$$ \mathrm{k} \propto \frac{1}{\mathrm{~L}} $$
This means that the product of the spring constant and the length is a constant for a given spring:
$$ \mathrm{k} \cdot \mathrm{~L}=	ext { Constant } $$
The force constant of original spring is $\mathrm{k}_{\mathrm{o}}=15 \mathrm{~N} / \mathrm{m}$.
Let the original length be L .
The spring is cut into two pieces with a length ratio of $1: 3$.
$$ x+3 x=L \Rightarrow x=\frac{L}{4} $$
Hence, the length of smaller spring is $L_1=x=\frac{L}{4}$ and the length of larger spring is $L_2=3 x=\frac{3 L}{4}$.
Let $\mathrm{k}_1$ be the force constant of the smaller piece.
Using the inverse proportionality relationship:
$$ \mathrm{k}_{\mathrm{o}} \mathrm{~L}_{\mathrm{o}}=\mathrm{k}_1 \mathrm{~L}_1 $$
$$\Rightarrow $$ $$ 15 	imes \mathrm{L}=\mathrm{k}_1 	imes\left(\frac{\mathrm{L}}{4}ight) \Rightarrow \mathrm{k}_1=60 \mathrm{~N} / \mathrm{m} $$
Therefore, the force constant of smaller piece is $60 \mathrm{~N} / \mathrm{m}$.
Hence, the correct option is (C).

Answer

20

Answer

45

Answer

60

Answer

15

Oscillations

A spring of force constant 15 \mathrm{~N} / \mathrm{m} is cut into two pieces. If the ratio of their length is 1: 3, then the force constant of smaller piece is \_\_\_\_ \mathrm{N} / \mathrm{m}.

Choose the correct answer: