The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

M3 Jan 2007 - Simple Harmonic Motion Help

I am having trouble with part (d) of this question; here is the question:


A particle P of mass 0.25 kg is attached to one end of a light elastic string. The string has
natural length 0.8 m and modulus of elasticity lamda N. The other end of the string is attached
to a fixed point A. In its equilibrium position, P is 0.85 m vertically below A.

(a) Show that lamda = 39.2.

Hooke's law with the values I got lamda = 4g

The particle is now displaced to a point B, 0.95 m vertically below A, and released from rest.

(b) Prove that, while the string remains stretched, P moves with simple harmonic motion of period pi/7.

used F=ma => (g/4) - T = 0.25a

Solved to get x..= -196x, using w = 14, T = pi/7

(c) Calculate the speed of P at the instant when the string first becomes slack.

using v^2 = w^2 (a^2 - x^2), with a=0.1, x= 0.05, I got v = 1.21m/s.

The particle first comes to instantaneous rest at the point C.
(d) Find, to 3 significant figures, the time taken for P to move from B to C.

First I used time under which it acts under gravity with v=u+at, with v=0, u=1.21, a = g, and got T = 1.21/g.

How do I use x=asin(wt)? How do I work out what value of x to use in the expression?

The mark scheme doesn't help very much in terms of guidelines.

Also, for part c, I took the amplitude to be the extension from 0.85m to 0.95m, is this the correct method of working out amplitude from questions such as this?

Many Thanks
Ok,

So you movement under gravity is correct, so this is the time from when the string goes slack till the velocity goes to zero.

Now to get the time from it to move from the bottom to the slack position we can use the SHM equation.

As its starting at maximum amplitude we use x= Acos(wt). For x we know that it must go past the equilibrium position so we define that as a negative. It is hard to explain without a diagram and so if you don't understand I will try drawing you one. But if you just used +0.05 you would get the time for it to move to the point below the equilibrium position instead of above.

Now for the Maths.

x = -0.05 A = 0.1 w = 14

So -0.05 = 0.1cos(14t)
-0.5 = cos(14t)
cos^-1(-0.5) = 14t
Therefore 14t = (2/3)Pi
t = 0.1496

So then you can add your two times together and get the answer.

Spoiler

(edited 9 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you