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western50 · Apr 17, 2011, 03:50 PM

A block with mass m = 7.7 kg is attached to two springs with spring constants kleft = 28 N/m and kright = 56 N/m. The block is pulled a distance x = 0.25 m to the left of its equilibrium position and released from rest. The time the block to return to equilibrium for the first time is 0.476s, and speed of the block as it passes through the equilibrium position is 0.825m/s

Question: 1. Where is the block located, relative to equilibrium, at a time 1.14 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)?
how to write the position in a function of time? I got 0.145m and it is the wrong answer

2. What is the net force on the block at this time 1.14 s? (a negative force is to the left; a positive force is to the right)? I got -12.18 N, is that right?

Unknown008 · Apr 18, 2011, 09:08 AM

Could you post your work please? How did you get 0.145 m?

Then, the easiest way to get F is to first get a,

$a = -\omega^2 x$

$F =ma$

western50 · Apr 18, 2011, 03:58 PM

so A=Vmax*Sqrt(m/K)=0.249, and then y(t)=-0.249sin(w*t), where w=3.303, and t=1.14, that's how I got 0.145m

Unknown008 · Apr 19, 2011, 08:36 AM

Um... you don't need to use the formula $A = v_{max}\sqrt{\frac{m}{k}}$ since when the mass is pulled on the left, that's the maximum potential energy it gets and hence, cannot go further than 0.25 m to the left (which is the amplitude).

Then, here you have to use:

$y(t) = A\cos(\omega t)$

or

$y(t) = A\sin$\omega t + \frac{\pi}{4}$$

Because you at t = 0, the mass is not at displacement 0 m, but has a 'lead' (or path difference) or pi/4.