Question

Two springs of force constants $K_1$ and $K_2$ are connected to a mass m placed on a horizontal frictionless surface as shown in figure. (a) and (b). what is the ratio of the time periods of horizontal oscillation in cases (a) and (b) if $K_1=K_2$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 2
• D. 4

Answer 1

The correct option is C 2

In case (a) the springs are connected in series. Therefore, the spring constant of the combination is given by
$\frac{1}{k^{\prime }}=\frac{1}{k_1}+\frac{1}{k_2}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}(\because k_1=k_2=k)$ Therefore, for case(a),
$k^{\prime }=\frac{k}{2}$
$T_a=2\pi \sqrt{\frac{m}{k^{\prime }}}=2\pi \sqrt{\frac{2m}{k}}$
In case (b) the spring constant of the combination is
$k^{\prime \prime }=k_1+k_2=2k$
$\therefore T_b=2\pi \sqrt{\frac{m}{k^{\prime \prime }}}=2\pi \sqrt{\frac{m}{2k}}$
Thus $\frac{T_a}{T_b}=2$
Hence the correct choice is (c)