To find the amplitude of a mass on an elastic string

In summary, Mitch is having problems with calculating the time it takes for a mass to move from one point to another. He uses x = A sin (n t + φ), where n is the square root of 4g/L, to find t = 0. He then uses t = 1/(2) to find t = 0.5 and t = 1 to find t = 3. He then uses t = 3 to find x = -\frac l 4.
  • #1
gnits
137
46
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:

18.JPG


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.
 
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  • #2
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)##.
 
Last edited:
  • #3
PS I just checked through this and get the book answer.
 
  • #4
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.
 
  • #5
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!
 
  • #6
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.
 
  • #7
gnits said:
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.
 
  • #8
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.
 

Related to To find the amplitude of a mass on an elastic string

1. What is the definition of amplitude in the context of a mass on an elastic string?

In physics, amplitude refers to the maximum displacement or distance of an object or wave from its equilibrium position. In the case of a mass on an elastic string, amplitude represents the maximum distance the mass moves away from its resting position when the string is stretched.

2. How is the amplitude of a mass on an elastic string measured?

The amplitude of a mass on an elastic string can be measured by using a ruler or measuring tape to determine the maximum distance the mass moves away from its equilibrium position when the string is stretched. This distance is then recorded as the amplitude.

3. What factors affect the amplitude of a mass on an elastic string?

The amplitude of a mass on an elastic string can be affected by several factors, including the tension of the string, the mass of the object, and the stiffness of the string. A higher tension, lighter mass, and stiffer string will result in a larger amplitude.

4. Why is it important to find the amplitude of a mass on an elastic string?

Finding the amplitude of a mass on an elastic string is important in understanding the behavior of the string and the relationship between the mass and the string's elasticity. It can also help in determining the energy and forces involved in the system.

5. Is there a mathematical formula for calculating the amplitude of a mass on an elastic string?

Yes, the amplitude of a mass on an elastic string can be calculated using the equation A = F/k, where A is the amplitude, F is the applied force, and k is the spring constant of the string. This formula assumes that the string is ideal and obeys Hooke's Law.

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