- #1

CAF123

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## Homework Statement

A mass m hangs in equilibrium at the lower end of a vertical spring of natural length a, extending the spring to be a length b.

1) Show that the frequency for small oscillations about the point of equilibrium is ##\omega = \sqrt{g/(b-a)}##

2) The top end of the spring is made to oscillate vertically with a displacement ##c\sin (nt)##, moving downwards at t=0. Show that as a result, the length of the spring L(t) obeys: $$\ddot{L} + \omega^2 L = \omega^2 b + cn^2 \sin (nt)$$

## The Attempt at a Solution

1)Write the eqn of motion: ##m\ddot{y} = -mg + k(y - a)##. In equilibrium, y = b, so ##mg = k(b-a) \Rightarrow k = mg/(b-a)##. Taking the period of a mass on a spring in the case of small oscillations, ##T = 2 \pi \sqrt{m/k}##, I get that ##T = 2 \pi \sqrt{(b-a)/g} \Rightarrow \omega = 2 \pi/T = 2 \pi \cdot \sqrt{g/(b-a)} \frac{1}{2 \pi}## which gives the result. Is this ok - am I allowed to simply quote the formula for the period of a mass on a spring? (I assumed yes, since it is given in the question that we are interested in small oscillations, so the formula is valid).

2) If it is made to oscillate vertically with that displacement then the spring must be under some sort of external driving force. I haven't been able to make much progress with this one. I know the displacement is given by y = y(t) = c sin (nt). This means the external driving force causing this is of the form ##mcn^2 sin(nt)##. So ##m\ddot{L} = mg - k(L - c \sin(nt)) - mcn^2 \sin(nt)##, but this does not resemble what I need to show.

Many thanks.