the body. Answer The angular frequency ω of the body = π s – and its time period T = s. At t = . s displacement = ( .
m = – . m Fig. . Answer Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig.
. . Under this situation the spring on the left side gets Fig. .
elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then, F = –k x (force exerted by the spring on the left side, trying to pull the mass towards the mean position) F = –k x (force exerted by the spring on the right side, trying to push the mass towards the mean position) The net force, F , acting on the mass is then given by, F = –2kx Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is, T = 2k ⊳ . ENERGY IN SIMPLE HARMONIC MOTION Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values.
In section . we have seen that the velocity of a particle executing SHM, is a periodic function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy ( K ) of such a particle, which is defined as mv K sin ( + ) A sin ( + ) = k A ( .
) is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position. Note, since the sign of v is immaterial in K , the period of