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actually i need help >.<
1) An oscillator consists of a block attached to a spring (k = 460 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.0875 m, v = -17.5 m/s, and a = -121 m/s2. Calculate (a) the frequency (in Hz) of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.
2) If the phase angle for a block-spring system in SHM is π/9 rad and the block's position is given by x = xmcos(ωt + φ), what is the ratio of the kinetic energy to the potential energy at time t = 0?
3) A thin uniform rod with mass m swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period T and an angular amplitude of φm (assume this angle is sufficiently small to allow for the use of the equations in this chapter). (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings? State your answers in terms of the given variables, using g and π when applicable.Edited by eagle 07 Mar `08, 1:55PM
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I think:
1) a)
Equation: a = -ω2x
So ω2 = -a / x = 1382.857
Thus ω = 2πf = 37.186787
f = 5.92 Hz
b) T = 2π sqrt (m/k)
But T (period) = 1/f
m = 0.332 kg (3 sf)
c) x = Acos(wt+f) -- (1)
v = dx/dt = -wAsin(wt+f) => Asin(wt+f) = -V / w -- (2)(1) ^2 + (2) ^2 = A^2 = x^2 + (-V/w )^2
So using the instant where x = 0.0875 and V = -17.5
A = 0.479 m
Someone confirm for me because no answer was posted...
And lazy to do Q2 and Q3 right now.
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3) A thin uniform rod with mass m swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period T and an angular amplitude of φm (assume this angle is sufficiently small to allow for the use of the equations in this chapter). (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings? State your answers in terms of the given variables, using g and π when applicable.
a) cannot remember if it is done like that...
let length be 2L, so CG of rod is at length L from the pivot
T = 2π sqrt (L/g)
Hence, L = (T/2π)^2 * gLength of rod = 2L = 2 * (T/2π)^2 * g
b) At amplitude, vertical distance of CG from pivot = L cos φm
Hence, PE gained from rest point = mgL (1 - cos φm)
KE = 0 at this pointAt max KE, PE = 0
By conservation of energy, max KE = mgL (1 - cos φm)
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