Question

As shown in figure a simple harmonic motion oscillator having identical four springs has time period.

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

Answer 1

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

In the given figure, the two springs above are in parallel with each other, giving an equivalent spring constant of $k+k=2k$.

Similarly the two springs below are also connected in series with each other, and giving an equivalent spring constant of $k+k=2k$.

These two springs are connected in series giving net spring constant for the system $=\frac{\left(2k\right)\left(2k\right)}{2k+2k}=k$

Hence the time period of the motion of system $=2\pi \sqrt{\frac{m}{k}}$