
#1
Apr1707, 11:09 AM

P: 4

The problem is the following:
a.) Obtain the equation of motion for the very small oscillations of a bead of mass m attached 1/5th of the way along a massless string of length 5l, which is under tension T. b.) Hence show that the angular frequency of oscillation is omega=sqrt(5T/4ml)  Now, i would start looking at this problem by looking at the problem by observing the following relation: Force = spring constant * displacement = mass * angular frequency of oscillation squared * displacement That is: F = sx = mx omega^2 Now, presumably force F can be substituted by tension T. This however is where i start to run into trouble. Normally, to calculate the equation of motion for a SHO i would start by calculating omega, and the amplitude and use that to calculate the equation of motion. However this question seems to want me to do the exact opposite. I also am not sure how to calculate omega as, although i am given the length of the string, i am given neither its spring constant, nor the amplitude of vibration of the bead. In fact the question does not even specify whether the bead is oscillating along the length of the string, in which case the string would behave as two springs, or perpendicularly. As you can see i am very lost and help to put me on the right track would be appreciated. Thanks Chopo 



#2
Apr1707, 06:02 PM

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Since the question doesn't say anything about the spring constant of the string, I think it means oscillations perpendicular to the length of the string.
Draw a picture with the string displaced by a distance "x" and find the force acting on the mass from the two parts of the string. You will find the force is proportional the the displacement, so it acts the same way as a spring. Assume the displacements are small, so siin theta = theta for the small angles. 



#3
Apr1707, 06:07 PM

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The bead is oscillating perpendicularly. Draw a picture of a small displacement and label an angle theta (say the angle to the end that's about 4 meters away). 'small displacement' is a key that you can use the approximation sin(theta)=theta. Calculate the restoring force (from the tension) and displacement. You can DERIVE the spring constant.




#4
Apr1807, 09:31 AM

P: 4

Simple Harmonic Oscillator
Ok, So for part a, obtaining the equation of motion, i guess i should start by looking at the generic equation of motion:
displacement = amplitude * cosine ( angular frequency of oscillation * time + phase shift ) x(t) = A cos( omega*t + phi) When x(t) = A, cos( om*t + phi) = 1 and om*t + phi = theta Now my next step could be wrong as i am not sure as to which side of the string angle theta refers. Arbitrarily, i have, therefore, drawn a triangle of angle theta, adjacent 4 L/5 and opposite side A. Here is my next leap of faith. Based on your hint that for small theta, theta ~ sin(theta), i would argue that for theta << 1 rad tan(theta) ~ theta. Hence amplitude A = 4 L tan(theta)/5 ~ 4 L*theta/5. At this point i need to find omega, so i guess i need to solve part b in order to solve part a. I know F = sx and omega=sqrt(s/m) So om = sqrt(F/(mx)) Where x is the displacement. Assuming F=T when x=A ~ 4 L*theta/5, i get: omega = sqrt(5*T/(4Lm*theta)). Which would be the correct answer if it where not for the theta term. I must have done something wrong because, if theta << 1, then theta > 0 and 1/theta > infinity. Thanks for the help so far, what am i doing wrong? Chopo 



#5
Apr1807, 09:37 AM

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To start with, which angle did you label theta? Your amplitude A is wrong  also the conclusion that F=T is way wrong.




#6
Apr1807, 10:22 AM

P: 4

The angle of a triangle of adjacent side 4L/5 and and opposite side A (amplitude)




#7
Apr1807, 01:01 PM

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I don't see any sides of length 4L/5.




#8
Apr1907, 07:20 AM

P: 4

The string is of length 5L and the bead is positioned 1/5 th of the way along the string. 4/5 ths of the way in the other direction. Hence 4/5 L




#9
Apr1907, 07:55 AM

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1/5th of the way along a string of length 5L is L. It splits the string into lengths L and 4L.



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