Question

A simple harmonic oscillator having four identical springs as shown in figure has time period (neglect the separation between the ends on the roof and the ends connected to the mass):

• A. $2\pi \sqrt{\frac{m}{4k}}$
• B. $2\pi \sqrt{\frac{m}{2k}}$
• C. $2\pi \sqrt{\frac{m}{k}}$
• D. $2\pi \sqrt{\frac{m}{8k}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{m}{k}}$


From figure, we can see that upper two springs $(k_1,k_2)$ and lower two springs $(k_3,k_4)$ are in parallel.
So,
$k_1+k_2=k+k=2k$, and $k_3+k_4=k+k=2k$
On simplifying the combination,

Again, the resultant of $(k_1,k_2)$ and resultant of $(k_3,k_4)$ will be in series.
So,
$\frac{1}{k_{eq}}=\frac{1}{2k}+\frac{1}{2k}$
$\therefore k_{eq}=k$
Hence, time period of oscillation for spring- block system is,
$T=2\pi \sqrt{\frac{m}{k_{eq}}}$
$\Rightarrow T=2\pi \sqrt{\frac{m}{k}}$