exercise-moving mass m-differential equations!

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evinda

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Hello!!! :D Could you help me at the following exercise??

Prove that moving mass m underlying the action of linear spring constant k, has the form y (t) = Asin (wt + f), where t is time and A, w, f fixed. Interpret the physical significance of these constants and determine their prices if at the time t = 0, the mass is removed y0 and velocity v0. If in addition the mass subject to outdoor force F (t) = \(\displaystyle F_ {0} sin (w_ {0} t)\), amplitude \(\displaystyle F_ {0} \) and cyclic frequency \(\displaystyle w_ {0}\), calculate the amplitude of motion and investigate the dependence of the circular frequency \(\displaystyle w_ {0} \).
 
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evinda said:
Hello!!! :D Could you help me at the following exercise??

Prove that moving mass m underlying the action of linear spring constant k, has the form y (t) = Asin (wt + f), where t is time and A, w, f fixed. Interpret the physical significance of these constants and determine their prices ...
Click to expand...

their prices? do you maybe mean their numeric values?
 
evinda said:
Hello!!! :D Could you help me at the following exercise??

Prove that moving mass m underlying the action of linear spring constant k, has the form y (t) = Asin (wt + f), where t is time and A, w, f fixed. Interpret the physical significance of these constants and determine their values if at the time t = 0, the mass is removed y0 and velocity v0. If in addition the mass subject to outdoor force F (t) = \(\displaystyle F_ {0} sin (w_ {0} t)\), amplitude \(\displaystyle F_ {0} \) and cyclic frequ ency \(\displaystyle w_ {0}\), calculate the amplitude of motion and investigate the dependence of the circular frequency \(\displaystyle w_ {0} \).
Click to expand...
Okay, that's a good statement of the problem. You say "help" but what kind of help do you want? That is, what do you know and what can you do with this?

If it were me I would say that the "linear spring constant k" meant that the force was "ky" where y is the extended length of the spring and, becauce "force equals mass times acceleration", the acceleration, which is the second derivative of y with respect to t, is \(\displaystyle \frac{d^2y}{dt^2}= -\frac{k}{m}x\). I would then solve that differential equation. Now, do you know how to solve a differential equation of that kind?
 
evinda said:
Yes,I meant their numeric values...I am sorry!! :shock:
Click to expand...

all this is asking you to do is to solve the equation of motion for a perfect spring system with spring constant k and mass m.

Apply Newton's law F=m d2x/dt2 to obtain the differential equation where x is the position of the mass.

This page has everything you could possibly want to know about this problem.
 
HallsofIvy said:
Okay, that's a good statement of the problem. You say "help" but what kind of help do you want? That is, what do you know and what can you do with this?

If it were me I would say that the "linear spring constant k" meant that the force was "ky" where y is the extended length of the spring and, becauce "force equals mass times acceleration", the acceleration, which is the second derivative of y with respect to t, is \(\displaystyle \frac{d^2y}{dt^2}= -\frac{k}{m}x\). I would then solve that differential equation. Now, do you know how to solve a differential equation of that kind?
Click to expand...

I solved this differential equation..My result is \(\displaystyle y(t)=c_{1}cos(wt)+c_{2}sin(wt) \). But how can I write it in the form \(\displaystyle y(t)=Asin(wt+f) \)?
 
evinda said:
I solved this differential equation..My result is \(\displaystyle y(t)=c_{1}cos(wt)+c_{2}sin(wt) \). But how can I write it in the form \(\displaystyle y(t)=Asin(wt+f) \)?
Click to expand...

You really should have seen how to convert back and forth between these by the time you are doing this differential eq but it goes like this. (A, f) is called phasor notation. A is the amplitude of the phasor, f is the phase of it.

A sin(wt + f) = A[sin(wt)cos(f) + cos(wt)sin(f)]

so c1 = A sin(f), c2 = A cos(f)

with a bit of algebra you can find that

A = sqrt(c12 + c22)

f = arctan(c1/c2) where you have to be careful choosing the actual quadrant of f by looking at the signs of c1 and c2
 
Ok!
And how can I find the amplitude of the motion knowing that the externally applied force is \(\displaystyle F(t)=F_{0}sin(w_{0}t) \)??
 
evinda said:
I solved this differential equation..My result is \(\displaystyle y(t)=c_{1}cos(wt)+c_{2}sin(wt) \). But how can I write it in the form \(\displaystyle y(t)=Asin(wt+f) \)?
Click to expand...
You should know that sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B). With \(\displaystyle y(t)= c_1 cos(\omega t)+ c_2 sin(\omega t)\), it is obvious to take \(\displaystyle B= \omega t\). Then we need A so that \(\displaystyle sin(A)= c_1\) and \(\displaystyle cos(t)= c_2\). In general we can't do that because we would have to have \(\displaystyle c_1^2+ c_2^2= sin^2(A)+ cos^2(A)= 1\) which isn't necessarily true. But we can take \(\displaystyle c= \sqrt{c_1^2+ c_2^2}\) and then look at \(\displaystyle C(\frac{c_1}{c}cos(B)+ \frac{c_2}{c}sin(B)\). Now, we can look for A such that \(\displaystyle sin(A)= \frac{c_1}{c}\) and \(\displaystyle cos(A)= \frac{c_1}{c}\). That is, take \(\displaystyle A= \arcsin\left(\frac{c_1}{\sqrt{c_1^2+c_2}}\right)\) so that \(\displaystyle c_1 cos(\omega t)+ c_2sin(\omega t)= \sqrt{c_1^2+ c_2^2}sin\left(\omega t+ \arcsin\left(\frac{c_1}{\sqrt{c_1^2+ c_2^2}}\right)\right)\).
 
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