Question 30

[sfarukhi1817]

I still dont understand why in this question delta x (change in equilibrium position) is 20cm and not 30 cm (70 - 40cm)

[jvuofm1709]

Please revise this question. The energy conservation law of .5Kx^2=.5mv^2 only holds true for HORIZONTAL oscillating springs. I will go over it below, but you can plug in the numbers to both scenarios yourself, and you will see that though MAXIMUM P.E. is conserved at all points, in a vertical spring the max K.E. will never equal the max Elastic potential.

To begin, in an ideal horizontal spring, the only Energy transfer is from elastic potential to kinetic. So max elastic potential=max kinetic potential. If mass is added to the spring and stretched the same distance...it will reach the SAME max K.E. at the equilibrium point as a lighter mass will. Mass will effect the period/frequency and velocity of oscillation, but never the total energy.

Now, in a vertical spring, the game changes. Now there are 3 different forms of energy being transferred, gravitational P.E., elastic P.E., and K.E. It is important to note, that a vertical spring does NOT oscillate about the unstretched length of spring, it will oscillate about the point where x=mg/k.

Now, the maximum velocity of the system is highest through the NEW equilibrium point (not when the elastic potential=0 as in a horizontal system). At this point, K.E., G.P.E., and E.P.E. all contribute to the total Energy, however K.E. is still maximized at this point:the center of oscillation.

This can be proven by examining the angular velocity of the system. So, in the original problem m=.5, and angular velocity(sqrt(k/m))=sqrt(50/.5)=10, the maximum amplitude=.2 m, and since angular velocity * max amplitude=max velocity: .2*10=2m/s. This value is pre-calculated in the problem. And the max K.E.= .5*.5*2^2=1 J

However, now let's attach .1Kg at the lowest amplitude as in problem 30. The total energy stays the same, since .5*k*x^2=.5*k*.3^2=2.25. This value is unaffected by mass. However, both the max G.P.E. and K.E. will be decreased. The G.P.E. is the easiest to visualize: the spring is accelerating a BIGGER mass with the SAME force...it won't go as high so it will have less max G.P.E.

To calculate max K.E. the first question to ask is: about what point will the new system oscillate?

the answer is calculated by using x=mg/k again. in this case x=.6*10/50=.12 m. So, since the new system has a heavier mass, the center of oscillation will be lowered to .12 m (below the unstretched length of the spring), as opposed to .1 m when the mass was only .5 Kg.

Now, the spring is still stretched by .3 meters, so the new max amplitude is merely how far the string is stretched passed its new oscillation point: .3-.12=.18m.

Next, we calculate the NEW maximum velocity of the system using sqrt(k/m)*max amplitude: sqrt(50/.6)*.18.=1.64316 m/s. So the NEW maximum K.E.=.5*m*v^2=.5*.6*1.64316^2=.81 J.

The max K.E. of the new system is lower than the original system. This difference can be exagerrated at longer lengths or mass differences, so please revise the problem and note that in a vertical spring, max K.E. does not ever equal max Elastic Potential.

[ljshi]

Why can't you use the equation:

a=-wx^2

where w=angular frequency (= sqrt(k/m))

When you use this formula, you get an answer of 2m/s^2 at maximum displacement.

[jellywing_2058]

Here is the solution to this problem:

M g = k x
0.5 x 10 = k x 0.10
k = 5/0.1
k = 50 N/m Spring Constant
A = 0.20 m
Vm = w A Maximum Velocity
2 = w x 0.20
w = 10 rad/s Angular frequency
0.5 M x Vm2 = 0.5 x 0.5 x 22 = 1 J

w’ = √ (k/M’) = √ (50/0.6) = √ (250/3)
The new angular frequency for the new mass M’ = 0.5 + 0.1 = 0.6 kg

Vm’ = w’ A = √ (250/ 3) x 0.2 The new Max. Velocity

New Maximum Kinetic Energy
KE = 0.5 x M’ x Vm’ = 0.5 x 0.6 x (250/ 3) x 0.04
KE = 0.004 x 250 = 1 J
Choice A.

[dnpgr16513]

Why isn't X .3m?? This questions hasn't been answered and I'm wondering if anyone knows the reason why.

Thanks.

[t.ahmadi82819]

there is a new equilibrium point for a vertical spring--that accounts for the length of the spring with the mass attached (50 cm). The displacement is 20 cm since that is how much it is stretched from equilibrium. Look at the previous posts they helped me

[t.ahmadi82819]

Yes,
you can use the formula for acceleration of an object in simple harmonic motion. You are supposed to square the angular velocity (w) and not displacement.
a(t)= -w^2 * x(t)= (10^2)(0.02) = 20 m/s