A particle of mass 5 kg is suspended from a fixed point by a light elastic stringwhich hangs vertically. The elastic constant of the string is 500 N/m.The mass is pulled down a vertical distance of 20 cm from the equilibriumposition and is then released from rest.(i) Show that the particle moves with simple harmonic motion.
- Thread Starter
- 18-08-2016 22:35
- 18-08-2016 23:50
Simple harmonic motion just requires a restoring force which is proportional to the displacement from the equilibrium position and some initial displacement.
- Thread Starter
- 19-08-2016 12:59
what i tryed to do was find the force down and the force up find the resultant force and let it equal to F=5a and then that would prove it but to do this when i am finding the force in the string i need the natural length of the string but it is not given in the question. to find the force up i use F=k(length-natural length)
- 19-08-2016 20:33
Before you start any work, contemplate the question. You've got a mass hanging on essentially a spring. The mass is 5kg... does that make any difference to the simple harmonic motion? Does a mass that is 5kg show SMH and a mass that's 4kg or 6kg not? No! The numbers are red herrings here.
To show something follows SHM, you have some common steps:
1. Find an expression for the force acting on the particle
2. Substitute this in for the "F" in Newton's 2nd law (F=ma)
3. Show that x = sin(wt) is a solution of this equation
Spoiler:ShowIn this case:
1. F = -kx
2. -kx = ma
3. x = sin(wt); v = w*cos(wt); a = -w^2sin(wt);
sub in: -k*sin(wt) = m*(-w^2*sin(wt));
divide: -k = -m*w^2
This equation is solved when w = sqrt(k/m). So: x = sin(wt) is a solution, and we have shown SHM.Last edited by mik1a; 19-08-2016 at 20:36.