- #1

Sam

- 14

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a 0.28 kg mass is suspended on a spring which stretches a distance of 5.6 cm. The mass is then pulled down an additional distance of 14 cm and released. What is the displacement from the equilibrium position with the mass attached (in cm) after 0.4 seconds? Take up to be positive and use g = 9.81 m/s^2

Ok, I have done the following and it's not correct:

This is how I tried to solve...

1. Find k, the spring constant F = kx (Hooke's Law).

Since F = mg, then k = mg/x = 4.905000

2. Find the natural frequency, omega = (k/m)^1/2.

Omega = 13.23550 radians/sec = 2.106495 Hz.

Since energy in the system is conserved, the amplitude of the oscillation is just +/-14 cm centered about the resting deflection of -5.6 cm.

At t = 0 the things starts out fully deflected (i.e., at the peak of the sinusoidal curve), so the equation of the motion would be:

x(t) = -0.056 - 0.14*cos(1.323550*t)

x(0.4) = -0.1329355 m

What did I do wrong? Please help.

Thank you!