# Find the time period for pulley spring system [closed] What I tried to do was first consider this system at equilibrium, assuming the pulley to be a disc of radius $$'r'$$ and I calculated the extensions of the 2 springs as $$e(=mg/k)$$ for lower spring and $$3e$$ for upper spring.

Then I decided to write the total energy of the system and equate it to a constant since energy is conserved in SHM.

$$\frac{1}{2}mv^2+\frac{1}{2}m\big(\frac{v}{2}\big)^2+\frac{1}{2}I\omega^2+\frac{1}{2}k\big(\frac{x}{2}+3e\big)^2+\frac{1}{2}k\big(x+e\big)^2-mgx-mg\frac{x}{2}=k$$

Where,

$$m$$= mass of pulley and block

$$k$$= spring constant

$$v$$= speed of mass

$$I$$= moment of inertia of pulley about centre

$$\omega$$= angular speed of pulley

Then I differentiate this energy and equate it to $$0$$ to obtain an expression of the form $$\frac{d^2x}{dt^2}=-\omega^2x$$

But I am facing 2 issues: 1) The gravitational potential energy terms are not cancelling out. Does this mean that our motion is occurring wrt to some other origin?

2) Is the constraint of pulley and mass used correct? I thought that when the block is shifted down by $$x$$, the pulley shifts down by $$\frac{x}{2}$$ but my friend argued that this would mean that the lower spring does not extend? I can't quite see why and how spring should affect the constraint.

Any help would be appreciated.

• Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. Feb 20, 2020 at 8:15
• You need find a constrain reaction ship between both Springs. If you lower down lower spring which is attached to mass m by x distance then what happened to the string which is attached between pulley and roof. Feb 20, 2020 at 8:18
• @YuvrajSingh… The upper spring is elongated by x/2. Feb 20, 2020 at 8:22
• @JohnRennie But I have narrowed it down to 2 concepts where I have a problem... Feb 20, 2020 at 8:24
• OK then x=2mg/k. Feb 20, 2020 at 8:31