Mechanics 3 SHM Watch
A) A particle is moving in a straight line with SHM of period 2pi secs about a centre O. When the particle is 0.5m from O its speed in zero. Calculate the maximum speed of the particle and its speed when it is 0.25m from O.
A piston of mass 1.5kg is moving with shm inside a cylinder. It performs 20 complete oscilations per minute and the distance between the extreme points of its motion is 2m. Calculate the maximum value of the kinetic energy possesed by the piston during its motion, stating clearly any assumptions you have made about the piston in order to calculate this.
Thanks for any help
v=0 at x=0.5m. Hence the amplitude is 0.5m.
Vmax at x=0:
v^2 = w^2(a^2 - x^2)
v^2 = 1(.5^2 - 0)
v = 0.5ms^-1
V at x=0.25
v^2 = 1(.5^2 - 0.25^2)
1 oscillation per 3 sec.
Period = 3sec
2pi/w = 3
w = 2pi/3
Amplitude = 2/2 = 1m.
Ke max = 0.5m(vmax)^2
(Vmax)^2 = (2pi/3)^2[1 - 0^2]
Vmax = 2pi/3
Ke max = 0.75[2pi/3]^2
Assumptions: There are no resistances to the movement of the piston and it is small so can be modelled as a particle.
When speed is zero, distance is maximum, so amplitude(A)=0.5 m
Maximum speed is when the particle is at the center, i.e. when x=0.
v²=w²(A²-x²), when x=0 => v²=w²A², so:
Max speed = wA = 0.5 m/s
Speed @ x=0.25 can be found using:
1 oscillation is 2pi, 1 minute is 60 s => 20 osc/min = 20(2pi/60) = 2pi/3 rad/s = w
Extereme points of motion are 2m apart, so amplitude = 1m.
Maximum kinetic energy occurs when speed is maximum, so:
max speed = wA = 2pi/3 m/s
max KE = 0.5mv² = 0.5(1.5)(2pi/3)² =~ 3.29 J
Piston was modelled as a particle.