To find the amplitude of a mass on an elastic string

[gnits]

Homework Statement

To find the amplitude of a mass oscillating on an elastic string

Relevant Equations

F=ma

Could I please ask for help with the following:

It's the final part I am having problems with.

So at this point we know we have SHM of amplitude (5/4)L

Now, using x = a cos(nt) ----- where we now know that n = sqrt(4g/L)

we can find the time at which x has any value.

So for exmaple, when x first equals (5/4)L we have:

(5/4)L = (5/4)L * cos( sqrt(4g/L) t ) which leads to t = 0, as expected.

Now, to find when x first arrives at C is to find when x first has a value of -(5/4)L giving:

-(5/4)L = (5/4)L * cos( sqrt(4g/L) * t1 ) which leads to:

t1 = (PI/4)*sqrt(L/g)

Similarly, to find when x first equals L/4 (i.e. when the mass is first at B) we can use:

L/4 = (5/4)L * cos( sqrt(4g/L) * t2) which leads to:

t2 = sqrt(L/g) * arccos(1/5)

and so time taken to move from B to C is:

t1 - t2 = (1/2) * ((PI/2) - arccos(1/5)) * sqrt(L/g)

which is not quite the desired answer.

Thanks for any help,
Mitch.

 

[PeroK]

At $t =0$ the mass is at $B$, which is not at an amplitude of $5l/4$. I think you have to calculate where in the SHM cycle the motion starts.

Hint: you should be looking at motion of the form $A\sin(\omega t + \phi)$.

 

[PeroK]

PS I just checked through this and get the book answer.

 

[gnits]

Thanks for your suggestion, that really helped. So I've almost get there but I have sign out, can you spot my error?

So I use $x=A\sin(\omega t + \phi)$

And we have that $A=\frac{5l}{4}$ and $\omega=\sqrt{\frac{4g}{l}}$

In order to have $x=\frac{l}{4}$ when t = 0 we need to have $\phi=arcsin(\frac{1}{5})$

So to find time to C we need to solve:

$sin(\sqrt{\frac{4g}{l}}t+arcsin(\frac{1}{5}))=1$

which leads to:

$t=\frac{1}{2}(\frac{\pi}{2}-arcsin(\frac{1}{5}))*\sqrt{\frac{l}{g}}$

and I have a minus instead of a plus. Thanks for any help.

 

[PeroK]

If $x(0) = \frac l 4$, then what is $x$ when the mass is at $C$?

In fact, perhaps think more carefully about taking $x(0) = \frac l 4$ in the first place!

 

[gnits]

I see, at C, $x=-\frac{5l}{4}$ (I had not used the negative)

This would lead to:

$sin(\sqrt{\frac{4g}{l}}t+arcsin(\frac{1}{5}))=-1$

and so

$\sqrt{\frac{4g}{l}}t+arcsin(\frac{1}{5})=-\frac{\pi}{2}$

and so:

$t=\frac{1}{2}(-\frac{\pi}{2}-arcsin(\frac{1}{5}))*\sqrt{\frac{l}{g}}$

Sorry if I'm missing something obvious, but I'm still off.

 

[PeroK]

I see, at C, $x=-\frac{5l}{4}$ (I had not used the negative)

This would lead to:

$sin(\sqrt{\frac{4g}{l}}t+arcsin(\frac{1}{5}))=-1$

and so

$\sqrt{\frac{4g}{l}}t+arcsin(\frac{1}{5})=-\frac{\pi}{2}$

and so:

$t=\frac{1}{2}(-\frac{\pi}{2}-arcsin(\frac{1}{5}))*\sqrt{\frac{l}{g}}$

Sorry if I'm missing something obvious, but I'm still off.

$x(0) = \frac l 4$ doesn't fit with taking the $\phi$ you got. If you differentiate your equation you find that the mass is moving away from equilibrium at $t =0$.

You could fix this by taking a different $\phi$ - note that more than one $\phi$ gives the same value of the trig functions. In any case, you need a different $\phi$ to get the correct position and velocity for the motion at $t = 0$.

Taking $x(0) = -\frac l 4$ and $C$ as $+\frac{5l}{4}$ is perhaps the simplest approach.

 

[gnits]

OK, that's very helpful indeed. I had not been looking at the direction of the velocity. Thanks very much for all your help. I'll work through it all agian to be sure to understand it fully.