Question

A body of mass 5.0 kg is suspended by a spring, which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately

a. y = .10sin 10t b. y = .10cos 10t c. y = .10cos (10t +.1)

d. y = .10sin (10t + 5) e. y = .05cos 10t

I'd like a second opinion on this one... As I've seen a few answers that say it's b "y= .10cos 10t". But many more that say it's e "y = .05cos 10t"

Answer 1

Solution:

First,

Find the spring constant,

Spring is stretches 10 cm when mass is attached stayed in equilibrium

Using Newton's law,

F_net = 0

mg = kx

k = mg/x = 5*9.80/0.1 = 490 N/m .

= k/m = 490/5 = 9.899 = 10 rad/s.

Then It is then displaced downward an additional 5.0 cm

A = 5 cm = 0.05 m

Finally,

Position as a function of time is approximately,

y(t) = ACost or ASInt

So, From the given options,

y(t) = 0.05 Cos 10t.

Ans: E

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