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Mass is subject to external force!!

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hello!!! :)
I have a question..I am given the following exercise:
Prove that the motion of a mass m on a linear spring with constant $k$, has the form $y (t) = Asin(wt+f)$ , where $t$ is the time and $A, w, f$ are constants. Interpret the physical meaning of the above constants and specify their values if for $t = 0, y(0)=y_{0}$ and $y'(0)=v_{0}$. If,in addition, the mass is subject to external force $F (t) = F_{0}sin (w_{0}t)$, where $F_{0}$ the amplitude and $w_{0}$ the cyclic frequency,calculate the amplitude of the motion and find its dependence from the cyclic frequency $w_{0}$.


I have shown that the motion of the mass has the form $y(t)= Asin(wt+f)$,where $A=\sqrt{\frac{v_{0}^{2}}{w^{2}}+y_{0}^{2}}, \text { where } w=\sqrt{\frac{k}{m}}$ and $f=arctan(\frac{y_{0}w}{v_{0}})$ .But,when the mass is subject to external force $F (t) = F_{0}sin (w_{0}t)$,do we get this differential equation: $y''+w^{2}y=\frac{F_{0}}{m}sin(w_{0}t)$ ,or am I wrong?If it is right,how can I find the amplitude of the motion? :confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For an undamped system with a sinusoidal forcing term, we could say it is governed by:

\(\displaystyle m\frac{d^2y}{dt^2}+ky=F_0\sin(\omega_0 t)\)

A general solution to this ODE is the sum of a particular solution and a general solution to the corresponding homogenous equation.

You have already found the form of the homogeneous solution. Can you now state the form of the particular solution, either by using a table or using the annihilator method? And then I suggest using the method of undetermined coefficients to determine the particular solution. At this point you can then use linear combination identities to determine the amplitude of the resulting motion and its dependence on the cyclic frequency.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
For an undamped system with a sinusoidal forcing term, we could say it is governed by:

\(\displaystyle m\frac{d^2y}{dt^2}+ky=F_0\sin(\omega_0 t)\)

A general solution to this ODE is the sum of a particular solution and a general solution to the corresponding homogenous equation.

You have already found the form of the homogeneous solution. Can you now state the form of the particular solution, either by using a table or using the annihilator method? And then I suggest using the method of undetermined coefficients to determine the particular solution. At this point you can then use linear combination identities to determine the amplitude of the resulting motion and its dependence on the cyclic frequency.
I have found the general solution: $y(t)=c_{1}cos(wt)+c_{2}sin(wt)+\frac{F_{0}}{m(w-w_{0}^{2})}sin(w_{0}t)$ where $c_{1}=y_{0} $ and $ c_{2}=\frac{v_{0}}{w}-\frac{F_{0}w_{0}}{mw(w-w_{0}^{2})}$ Is this right?? But how can I find the amplitude of the motion? :confused:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Do I have to write $y$ in the form $Asin(wt+f)$,or do I have to do something else?? (Thinking)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, we are given the IVP:

\(\displaystyle m\frac{d^2y}{dt^2}+ky=F_0\sin\left(\omega_0 t \right)\) where \(\displaystyle y(0)=y_0,\,y'(0)=v_0\)

The characteristic roots are:

\(\displaystyle r=\pm\sqrt{\frac{k}{m}}i\)

Hence, the homogeneous solution is:

\(\displaystyle y_h(t)=c_1\cos\left(\sqrt{\frac{k}{m}}t \right)+c_2\sin\left(\sqrt{\frac{k}{m}}t \right)\)

Applying a linear combination identity we can express this solution in the form:

\(\displaystyle y_h(t)=c_1\sin\left(\sqrt{\frac{k}{m}}t+c_2 \right)\)

Now, because of the form of the forcing term on the right, we may assume the particular solution must take the form:

\(\displaystyle y_p(t)=A\sin\left(\omega_0 t \right)+B\cos\left(\omega_0 t \right)\)

Differentiating twice, we find:

\(\displaystyle \frac{d^2}{dt^2}y_p(t)=-\omega_0^2y_p(t)\)

Substituting into the ODE, we find:

\(\displaystyle -m\omega_0^2\left(A\sin\left(\omega_0 t \right)+B\cos\left(\omega_0 t \right) \right)+k\left(A\sin\left(\omega_0 t \right)+B\cos\left(\omega_0 t \right) \right)=F_0\sin\left(\omega_0 t \right)\)

So that we may compare coefficients, we arrange this equation as follows:

\(\displaystyle A\left(k-m\omega_0^2 \right)\sin\left(\omega_0 t \right)+B\left(k-m\omega_0^2 \right)\cos\left(\omega_0 t \right)=F_0\sin\left(\omega_0 t \right)+0\cos\left(\omega_0 t \right)\)

Equating the coefficients, we obtain the system:

\(\displaystyle A\left(k-m\omega_0^2 \right)=F_0\implies A=\frac{F_0}{k-m\omega_0^2}\)

\(\displaystyle B\left(k-m\omega_0^2 \right)=0\)

Assuming \(\displaystyle k\ne m\omega_0^2\) we find $B=0$, and so our particular solution is:

\(\displaystyle y_p(t)=\frac{F_0}{k-m\omega_0^2}\sin\left(\omega_0 t \right)\)

And thus, by the principle of superposition, the general solution is given by:

\(\displaystyle y(t)=y_h(t)+y_p(t)\)

\(\displaystyle y(t)=c_1\sin\left(\sqrt{\frac{k}{m}}t+c_2 \right)+\frac{F_0}{k-m\omega_0^2}\sin\left(\omega_0 t \right)\)

Differentiating with respect to $t$, we find:

\(\displaystyle y'(t)=c_1\sqrt{\frac{k}{m}}\cos\left(\sqrt{\frac{k}{m}}t+c_2 \right)+\frac{F_0\omega_0}{k-m\omega_0^2}\cos\left(\omega_0 t \right)\)

Utilizing the given initial values, we find:

\(\displaystyle y(0)=c_1\sin\left(c_2 \right)=y_0\)

\(\displaystyle y'(0)=c_1\sqrt{\frac{k}{m}}\cos\left(c_2 \right)+\frac{F_0\omega_0}{k-m\omega_0^2}=v_0\)

The first equation gives us:

\(\displaystyle c_1=\frac{y_0}{\sin\left(c_2 \right)}\)

And so substituting for $c_1$ into the second equation, we obtain:

\(\displaystyle \frac{y_0}{\sin\left(c_2 \right)}\sqrt{\frac{k}{m}}\cos\left(c_2 \right)+\frac{F_0\omega_0}{k-m\omega_0^2}=v_0\)

We may arrange this as:

\(\displaystyle \tan\left(c_2 \right)=\frac{y_0\left(k-m\omega_0^2 \right)}{v_0\left(k-m\omega_0^2 \right)-F_0\omega_0}\sqrt{\frac{k}{m}}\)

And so we find:

\(\displaystyle c_2=\tan^{-1}\left(\frac{y_0\left(k-m\omega_0^2 \right)}{v_0\left(k-m\omega_0^2 \right)-F_0\omega_0}\sqrt{\frac{k}{m}} \right)\)

For simplicity, lets define:

\(\displaystyle \alpha\equiv\frac{y_0\left(k-m\omega_0^2 \right)}{v_0\left(k-m\omega_0^2 \right)-F_0\omega_0}\sqrt{\frac{k}{m}}\)

Hence:

\(\displaystyle c_2=\tan^{-1}(\alpha)\)

and then we find:

\(\displaystyle c_1=\frac{y_0\sqrt{\alpha^2+1}}{\alpha}\)

Thus, the solution satisfying the IVP is:

\(\displaystyle y(t)=\frac{y_0\sqrt{\alpha^2+1}}{\alpha}\sin\left(\sqrt{\frac{k}{m}}t+\tan^{-1}(\alpha) \right)+\frac{F_0}{k-m\omega_0^2}\sin\left(\omega_0 t \right)\)

We see we have the sum of two sinusoidal terms, with differing amplitudes, periods and phase shifts. I really can't think of a means to determine the amplitude of the resulting combination. Perhaps someone else can suggest a means of doing so, perhaps using Fourier analysis. :D