Physics help required

WoAiMeiMei SG

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.

eagle SG

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.

WoAiMeiMei SG

yay thanks eagle.. the first one is correct.

the last one is the hardest apparently >.<

eagle SG

Please post the answers if you have them (Homework Forum guidelines)

Either someone will help you, or I find some time tonight to do the rest... The first question is to help you get started....

WoAiMeiMei SG

yeah sorry i dont have answers ey...

WoAiMeiMei SG

hmm correction for q2 is:

If the phase angle for a block-spring system in SHM is π/9 rad and the block's position is given by x = x_mcos(ωt + φ), what is the ratio of the kinetic energy to the potential energy at time t = 0?

change: π/9 rad

eagle SG

2) At t=0

x = x_mcos(π/9)
PE = 1/2 k x^2 = 1/2 k [ x_mcos(π/9) ]^2

v = dx/dt = -ωx_msin(ωt + φ) = -ωx_msin(π/9)
KE = 1/2 m v^2 = 1/2 m [ωx_msin(π/9)] ^2
But ω = 2π / T and T = 2π sqrt (m/k)
hence, ω^2 = k/m

Thus, KE =1/2 k [ x_msin(π/9)] ^2

So, KE/PE = cos(π/9)^2 / sin(π/9)^2 = [ tan(π/9) ]^2

Ans: 0.132

eagle SG

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 * g

Length 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 point

At max KE, PE = 0
By conservation of energy, max KE = mgL (1 - cos φm)

WoAiMeiMei SG

if the answer for the last question is dimensionally correct, apparently it will produce these 2 answers:

L=0.64 m and Kmax=0.031 J