Question

Two springs are connected to a block of mass $M$ placed on a frictionless surface as shown below. If both the springs have a spring constant $k$, the frequency of oscillation of the block is

• A. $\frac{1}{2\pi }\sqrt{\frac{k}{M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{k}{2M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{2k}{M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{M}{k}}$

Answer 1

Answer: (B)

Let when the oscillating mass is at a distance $x$ towards right from its equilibrium position, the instantaneous extensions in the springs of force constants $k$, is

$x=x_1+x_2$
Since, the springs are in series the restoring force exerted by each spring on mass $m$ is same. Then
$F=-k{x}_1=-k{x}_2$
$\therefore x_1=-\frac{F}{k},x_2=-\frac{F}{k}$
and $x=x_1+x_2=-F\left(\frac{1}{k}+\frac{1}{k}\right)=\frac{-2F}{k}$
$\therefore F=-\frac{k}{2}x$
$\Rightarrow$ Effective force constant is $\frac{k}{2}$.
Hence, frequency of oscillation is
$n=\frac{1}{2\pi }\sqrt{\frac{k}{2M}}$