Question

The springs in fig. $A$ and $B$ are identical but the length of spring $A$ is three times the length of each spring in $B$. The ratio of period $T_A/T_B$ is

• A. $\sqrt{3}$
• B. $1/3$
• C. $3$
• D. $1/\sqrt{3}$

Answer 1

The correct option is C $3$

Length of each spring in $B$ is $L$ and spring constant is $K$
$\therefore K_{\text{eq}}=3K$
$\because K\propto \frac{1}{L}\Rightarrow L\propto \frac{1}{K}$
For $B:L_B\propto \frac{1}{K_B};(K_{\text{eq}}{)}_B=3K$
For $A:$ Length of $A$ is three times each of that in $B$.

$K_A\propto \frac{1}{L_A}\propto \frac{1}{3L_B}\propto \frac{K}{3}$
$\therefore (K_{\text{eq}}{)}_A=\frac{K}{3}$
$(K_{eq}{)}_A=K/3,K_B=3K\because \frac{T_A}{T_B}=\sqrt{\frac{K_B}{K_A}}\Rightarrow \frac{T_A}{T_B}=\sqrt{\frac{K_B}{K_A}}=\frac{3}{1}$