A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal support. The student is asked to find the spring constant, k. After suspending a mass of 255.0 g from the spring, the student notices the spring is displaced 47.5 cm from its previous equilibrium. With this information, calculate the spring constant. spring constant: N/m When the spring, with the attached 255.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calcula the period of oscillation, T, neglecting the mass of the spring itself.

Hooke's law states that the restoring force applied by a spring is proportional to the displacement of the spring and opposite in direction. The energy of a particle in SHM changes periodically from P.E to K.E. Both potential and kinetic energy depends upon the square of the displacement.(1) From Hooke's law- ⇒F=kx⇒mg=kx⇒k=mgx⇒k=0.255 kg×9.8 m/s20.475 m⇒k=5.26 N/m (2) Time period of oscillation is given by-T=2πmk ⇒T=2π0.255 kg5.26 N/m⇒T=1.38 s (3) For displacement of 35.6 cm- Potential energy is given as-U=12kx2=12×5.26×(0.356)2 =0.334 J Since 35.6 cm is the maximum stretch from the equilibrium position, i.e., 47.5cm, so at 35.6cm all of the energy is in the form of potential energy. Kinetic energy is zero at this point.⇒Velocity at 35.6cm = 0 m/s So, total energy of the spring-mass system is 0.334 J. For displacement of 26.1cm- Potential energy is given as-U'=12kx'2=12×5.26×(0.261)2 =0.180 J Since total energy of the system is constant and is equal to 0.334 J Therefore, Kinetic energy at 26.1cm is given by-Total energy(T)=Kinetic energy (K')+potential energy (U')⇒K'=T-U'⇒K'=0.334 J-0.180 J⇒K'=0.154 J Now,∵K'=12mv'2⇒v'=2K'm⇒v'=2×0.154 0.255 kg⇒v'=1.10 m/s Conclusion: (1) Spring constant is 5.26 N/m(2) Time period of oscillation is 1.38 s(3) At 35.6cm velocity is zero and potential energy is 0.334 J At 26.1cm velocity is 1.10 m/s and potential energy is 0.180 J