Question

A 318 g object is attached to a spring and executes simple harmonic motion with a period of 0.270 s. If the total energy of the system is 6.29 J.

(a) Find the maximum speed of the object. (------m/s ?)

(b) Find the force constant of the spring. (--------N/m ?)

(c) Find the amplitude of the motion. (------- m ?)

Answer 1

Okay, when the velocity is a max, all the energy of the system will be kinetic energy. This is because the max velocity occurs at x = 0, where x is the extension of the spring. As there is no extension of the spring, the energy stored in the spring is also zero.

The kinetic energy of the spring will be equal to 6.29 J. putting in 0.318 kg (318 gm) for the mass.

(a) 6.29 J = mv²/2 =) v = 6.29 m/s

(c) To find the spring constant, use the equation omega = 2pi/T, where T is the period, and omega is the angular velocity. Subbing 0.27 s for T and solving for omega gives 23.27 rad/s. Then using the equation v = r*omega, subbing in v = 6.29 m/s and omega = 23.27 rad/s, you get r = 0.27 m.

(b)When the spring is at max extension, the velocity is zero and therefore kinetic energy of the system is zero. The extension of the spring (x) is the same as the amplitude of motion and can be used to calculate the force constant, using the equation E = 1/2kx^2. Setting E = 6.29 J and x = 0.27, solving this for k gives, k = 172.56 N/m.

answers:

(a) 6.29 m/s

(b) 172.56 N/m

(c) 0.27 m