- #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:

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.

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.