A 0.500kg mass oscillates in simple harmonic motion at the end of a spring of co

Question

A 0.500kg mass oscillates in simple harmonic motion at the end of a spring of constant 150N/m. The total energy of the system is 2.50J. 1) What is the amplitude of the motion (in m)? 2) What is the maximum speed of the mass (in m/s)?


and...

A 0.500kg mass is attached to a spring of constant 150N/m. A driving force F(t) = ( 12.0N) cos(wt) is applied to the mass, and the damping coefficient b is 6.00Ns/m. What is the amplitude (in cm) of the steady-state motion if w is equal to half of the natural frequency w0 of the system?

Explanation / Answer

Mass of the block m = 0.5 Kg

Spring constant K = 150 N/m

Total energy E = 2.50 J

Angular speed = (K/m) = 17.32 rad/s

(a)

Total energy E = (1/2)KA^2   where A is amplitude

            2.50 = 0.5 * 150 * A^2

               A = 0.182 m

(b)

Maximum speed v = A = 0.182 * 17.32 = 3.162 m/s

NOTE : Only one question per one post,please post the second question seperately.Sorry.

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