Question

A uniform spring has certain mass suspended from it and its period for vertical oscillation is $T_1$. The spring is now cut into two parts having the lengths in the ratio of 1 : 2 and a mass m is suspended with these two springs as shown in the figure. Now, time period is $T_2$. The ratio $T_1/T_2$ is:

• A. 1
• B. $sin\theta$
• C. $\sqrt{\frac{2}{9}}$
• D. zero

Answer 1

The correct option is C $\sqrt{\frac{2}{9}}$

Let k be initial force constant of spring, $k_1$ and $k_2$ be force constant of new springs$.$

We can derive$,$

$l_1/l_2=2=>x_1/x_2=>2=>k_1/k_2=1/2$ ----------$\left(1\right)$

As initially these lengths were in series$-$

$1/k=1/k1+1/k_2$

$1/k=1/k1+1/2k_1$

Solving this$,$

$k_1=3k{/}_2$

$k_2=3k$

When these two strings are connected in parallel$,$

$k^{\prime }=k_1+k_2$

$k^{\prime }=3k/2+3k$

$k^{\prime }=9k/2$

Time period is$:$

$T^{\prime }/T=\surd \left(k/k^{\prime }\right)$

$T^{\prime }/T=\surd \left(2/9\right)$

hence,

option $\left(C\right)$ is correct answer.