# Spring extensions

A certain spring has attached to it a mass of 25 units: on increasing the load by 6 units it extends 2.5 cm.

a) What is the time of oscillation under the original load?

b) What will be the velocity and acceleration when it is midway between its lowest and mean positions if it is loaded as at first, pulled down 5 cm and let go?

I am guessing the spring is hung vertically. Do I work out the modulus of elasticity first? I guess the equations $\omega=\sqrt{k/m}$ will be helpful.

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Let's use SI units so mass is in kilograms and length is in meters. Let the original mass be $m_1 = 25 \,\mathrm{kg}$ and the final mass be $m_2 = m_1 +\Delta m$ where $\Delta m = 6 \,\mathrm{kg}$. Let the original extension of the spring be $x_1$ and the final extension be $x_1 + \Delta x_1$ where $\Delta x = 2.5\, \mathrm{cm}$. We can determine the spring constant by noting that in each case, the gravitational force of the mass on the spring must balance the spring force pulling up; this gives two equations:

$m_1g = k(x_1 - x_\mathrm{eq})$

$m_2g = k(x_2 - x_\mathrm{eq})$

so that subtracting the first equation from the second gives

$(m_2 - m_1)g = k(x_2-x_1)$

which given the notation above gives

$\Delta m g = k \Delta x$

so the spring constant is

$k = \frac{\Delta m}{\Delta x} g = \frac{6\,\mathrm{kg}}{2.5\,\mathrm{cm}} g$

With this in hand, you can indeed compute the angular frequency for the original load $m_1$ using the formula you wrote. This in turn will give you the period, which is what I assume part (a) is asking for. I'll let you think about part (b) since this sounds like a homework question to me.

Let me know of any typos.

Cheers!

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Just remember not to give away too much when you're answering a homework question ;-) – David Z Jan 25 '13 at 0:05
@DavidZaslavsky Yea; my apologies. I'll make my comments more general next time. – joshphysics Jan 25 '13 at 0:21
So the time period is $\omega = \sqrt{k/m_1}=\sqrt{12/(5gm_1)} 2\pi f=\sqrt{12/5gm_1} T=2\pi \sqrt{5gm_1/12}$ – bbr4in Jan 26 '13 at 13:43