Physics - SHM Question with Solution | TestHub

For system (a), there is a single spring of force constant $K$. The time period is:

$T_a=2\pi \sqrt{\frac{m}{K}}$

For system (b), two identical springs are connected in series. The equivalent force constant $K_b$ is given by:

$\frac{1}{K_b}=\frac{1}{K}+\frac{1}{K}=\frac{2}{K}\Rightarrow K_b=\frac{K}{2}$

The time period for system (b) is:

$T_b=2\pi \sqrt{\frac{m}{K_b}}=2\pi \sqrt{\frac{m}{K/2}}=2\pi \sqrt{\frac{2m}{K}}=\sqrt{2}{T}_a$

For system (c), two identical springs are connected in parallel. The equivalent force constant $K_c$ is given by:

$K_c=K+K=2K$

The time period for system (c) is:

$T_c=2\pi \sqrt{\frac{m}{K_c}}=2\pi \sqrt{\frac{m}{2K}}=\frac{1}{\sqrt{2}}\left(2\pi \sqrt{\frac{m}{K}}\right)=\frac{T_a}{\sqrt{2}}$

Now, comparing $T_b$ and $T_c$:

$\frac{T_b}{T_c}=\frac{2\pi \sqrt{\frac{2m}{K}}}{2\pi \sqrt{\frac{m}{2K}}}=\sqrt{\frac{2}{1/2}}=\sqrt{4}=2$

$\Rightarrow T_b=2T_c$

Answer: ${\mathrm{T}}_{\mathrm{b}}=2{\mathrm{T}}_{\mathrm{c}}$