Question

The springs in figure $A$ and $B$ are identical but length of spring in $A$ is three times than length of each spring in $B$. The ratio of period $T_A/T_B$ is:-

• A. $\surd 3$
• B. $1/3$
• C. $3$
• D. $1/\surd 3$

Answer 1

The correct option is C $3$

Spring constant $\alpha \frac{1}{Length}$

Let spring constant of spring A=k,then spring constant of each spring in B =3k

Now equivalent spring constant of B $=k_{eq}=3\left(3k\right)=9k$

Now, $T_A=2\pi \sqrt{\frac{m}{k}}$

$T_B=2\pi \sqrt{\frac{m}{9k}}$

$\frac{T_A}{T_B}=3$