Physics - Practical physics Question with Solution | TestHub

From the equilibrium position if upper spring elongates by $x$, elongation in the lower spring is $\frac{x}{2}\epsilon$ net displacement of block in the down ward defection is $\frac{x}{2}+2x=\frac{5}{2}x$ and incase in tension in the string connected to mass m is $\frac{kx}{2}$, it is the restoring force on blocks $m{\omega }^2\frac{5x}{2}=k\frac{x}{2}$ ${\omega }^2=\frac{k}{5m}$ or $T=2\pi \sqrt{\frac{5m}{k}}$

B) In this case if pulley moves down by ' $x$ ' from equilibrium, block displaced drum by $2x$ and incase in spring force in each spring is $\frac{kx}{2}$ incase in tension an the block is $\frac{kx}{4}$ So $\frac{kx}{4}=m{\omega }^{2}2x$ or ${\omega }^2=\frac{k}{8m}$ $T=2\pi \sqrt{\frac{8m}{k}}$

C) If pulley moves dawn by ' $x$ ' elongation of each spring dues the equilibrium is $x$. And incase in tension an the block is 2 kx so $2km=m{\omega }^{2}x$ or $T=25\sqrt{\frac{m}{2k}}$

D) If $x$ is the additional elongation in the upper spring, lower spring elongates by $2x$ and set displacement of block is $2x+\frac{x}{2}=\frac{5x}{2}$ change in tension an the block is $2kx$ so $2kx={\omega }^2=\frac{4k}{5m}$