Question

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 1.9 m/s. Determine the amplitude A of the motion.

Answer 1

the amplitude(x) of the motion = 0.078 m

Step-by-step explanation:

the maximum kinetic energy occurs at maximum speed, that is:

Kmax = 1/2×m×(v^2) = 1/2×(0.20)×(1.9)^2 = 0.361 J

then, in a simple harmonic motion, energy is conserved, the sum of kinetic energy and potential energy is always constant:

then, the maximum potential energy in the spring is given by:

Umax = 1/2×k×(x^2) , where x is the amplitude of the oscillation.

= 1/2×(120)×(x^2)

then by conservation of energy:

Kmax = Umax

0.361 J = 60×x^2

x^2 = 6.02×10^-3

x = 0.078 m

Therefore, the amplitude of the motion is 0.078 m.