# Spring Pendulum

A point mass suspended from a massless spring or placed on a frictionless horizontal plane attached with spring (Fig. 7) constitutes a linear harmonic spring pendulum

**Fig. 7**

Time period,

and frequency,

*Important Points*- Time period of a spring pendulum depends on the mass suspended
- The time period depends on the force constant
*k*of the spring - Time of a spring pendulum is independent of acceleration due to gravity. That is why a clock based on spring pendulum will keep proper time every where on a hill or moon or in a satellite and time period of a spring pendulum will not change inside a liquid if damping effects are neglected.
- If two masses of mass
*m*_{1}and*m*_{2}are connected by a spring and made to oscillate on horizontal surface, the reduced mass*m*is given by_{r}**Fig. 8** - If a spring pendulum oscillating in a vertical plane is made to oscillate on a horizontal surface, (or on inclined plane) time period will remain unchanged. However, equilibrium position for a spring in a horizontal plain is the position of natural length of spring as weight is balanced by reaction. While in case of vertical motion, equilibrium position will be
*L*+*y*_{0}with*ky*_{0}=*mg*.**Fig. 9** - If the stretch in a vertically loaded spring is
*y*_{0}, then for equilibrium of mass*m*,*ky*_{0 }=*mg,*i.e., .*g*because along with*g*,*y*_{o}will also change in such a way that remains constant. - The spring constant
*k*is inversely proportional to the spring length. - A spring of length
*l*is cut into two pieces of length*l*_{1}and*l*_{2}such that*l*_{1}=*nl*_{2}.*k*, then spring constant of first part,*k*_{2}= (*n*+ 1)*k**,*

*Important Facts and Formulae*- The pendulum clock runs slow due to increase in its time period whereas it becomes fast due to decrease in time period.
- If infinite spring with force constant
*k*, 2*k*, 4*k*, 8*k*, â€¦, respectively, are connected in series. The effective force constant of the spring will be*k*/2. - If
*y*_{1}=*a*sin*Ï‰t*and*y*_{2}=*b*cos*Ï‰t*are two SHM then by the superimposition of these two SHM we get*y*=*a*sin*Ï‰t*+*b*cos*Ï‰t*=*A*sin (*Ï‰t*+*Ï†*)*Ï†*= tan^{â€“1}(*b*/*a*) - If a particle performs SHM whose velocity is
*v*_{1 }at*x*_{1}distance from mean position and velocity*v*_{2}at distance*x*_{2}, then