Benny
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I'm stuck on question in where I need to set up a system of equations, can someone please help me with it? There are various things in this question which I just cannot figure out so any assistance would be great.
Q. A light string stretched to tension T has its ends fixed at a distance 9L apart. Masses 2m, m are attached to the string distances L, 5L respectively from one of its fixed ends. The system undergoes transverse vibrations.
Show that the equations of motion are
<br /> 8\mathop x\limits^{ \bullet \bullet } + n^2 \left( {5x - y} \right) = 0<br />
<br /> 4\mathop y\limits^{ \bullet \bullet } + n^2 \left( { - x + 2y} \right) = 0<br />
where n^2 = \frac{T}{{mL}}.
Note: Presumably they mean that the 2m and m masses weigh 2m kg and m kg respectively.
Sorry for the lack of a diagram but the situation is rather simple so I'll try to describe it. Basically, take a string, stretch it till it reaches a length of 9L and then fix each of its two ends so that the string is horizontal. This is defined by the question as the equilibrium configuration of the string.
The mass 2m is attached to the string at a distance L from the left end of the string while the mass m is attached to the string at a distance of 5L from the left end of the string.
When the system is set in motion the masses oscillate up and down. x(t) denotes the vertical displacement of the mass 2m while y(t) denotes the vertical displacement of the mass m.
This has been a rather frustrating problem for many reasons. I just can't figure out what needs to be done. I'm used to questions where the masses oscillate in the longitudinal direction. From what I can gather, there will probably be square roots all over the place. But here are my thoughts anyway.
When the string is in equilibrium (ie. when y(t) = x(t) = 0), the tension in the string is T as given in the question. In order to maintain static equilibrium of the two masses in the horizontal the tension in the string on either side of them must be equal to T. So the spring constant of the string is k = T/(9L)
When the masses start moving up and down the string is stretched further. So for example the distance between the mass 2m and the left most point increases from the original distance L to sqrt(L^2+x^2). But does this really mean anything in the context of the problem?
I can't get anywhere with this question nor can I figure out how n^2 could possibly be equal to T/(mL)...it just doesn't make sense because n^2 = T/(mL) implies that the spring constant of the string is T/L even though the length of the string is 9L and not L. Any helpful advice would be really good thanks.
Q. A light string stretched to tension T has its ends fixed at a distance 9L apart. Masses 2m, m are attached to the string distances L, 5L respectively from one of its fixed ends. The system undergoes transverse vibrations.
Show that the equations of motion are
<br /> 8\mathop x\limits^{ \bullet \bullet } + n^2 \left( {5x - y} \right) = 0<br />
<br /> 4\mathop y\limits^{ \bullet \bullet } + n^2 \left( { - x + 2y} \right) = 0<br />
where n^2 = \frac{T}{{mL}}.
Note: Presumably they mean that the 2m and m masses weigh 2m kg and m kg respectively.
Sorry for the lack of a diagram but the situation is rather simple so I'll try to describe it. Basically, take a string, stretch it till it reaches a length of 9L and then fix each of its two ends so that the string is horizontal. This is defined by the question as the equilibrium configuration of the string.
The mass 2m is attached to the string at a distance L from the left end of the string while the mass m is attached to the string at a distance of 5L from the left end of the string.
When the system is set in motion the masses oscillate up and down. x(t) denotes the vertical displacement of the mass 2m while y(t) denotes the vertical displacement of the mass m.
This has been a rather frustrating problem for many reasons. I just can't figure out what needs to be done. I'm used to questions where the masses oscillate in the longitudinal direction. From what I can gather, there will probably be square roots all over the place. But here are my thoughts anyway.
When the string is in equilibrium (ie. when y(t) = x(t) = 0), the tension in the string is T as given in the question. In order to maintain static equilibrium of the two masses in the horizontal the tension in the string on either side of them must be equal to T. So the spring constant of the string is k = T/(9L)
When the masses start moving up and down the string is stretched further. So for example the distance between the mass 2m and the left most point increases from the original distance L to sqrt(L^2+x^2). But does this really mean anything in the context of the problem?
I can't get anywhere with this question nor can I figure out how n^2 could possibly be equal to T/(mL)...it just doesn't make sense because n^2 = T/(mL) implies that the spring constant of the string is T/L even though the length of the string is 9L and not L. Any helpful advice would be really good thanks.