1.A particle attached to a spring with k = 54 N/m is undergoing simple harmonic
ID: 2300462 • Letter: 1
Question
1.A particle attached to a spring with k = 54 N/m is undergoing simple harmonic motion, and its position is described by the equation x = (6.0 m)cos(8.1t), with t measured in seconds.
(a) What is the mass of the particle?
(b) What is the period of the motion?
(c) What is the maximum speed of the particle?
(d) What is the maximum potential energy?
(e) What is the total energy?
2.A simple pendulum has a length L and a mass m. At its highest point, the pendulum mass is 0.25L above its lowest point (see figure below). What is the speed of the mass when it is at its lowest point? Express your answer in terms of m, L, and g.
3.A mass-on-a-spring oscillator has m = 6.7 kg and k = 274 N/m and is oscillating with an amplitude of 13 cm.
Find the original value and the new value of the maximum potential energy. (Enter your answers to two decimal places.)
PE original=? PE new=?
4. Use the exact values you enter in previous answer(s) to make later calculation(s).
Consider a Cavendish apparatus that employs a torsion fiber with
Explanation / Answer
1.
x = 6 cos (8.1*t)
Comparing it with standard eqn, x = a cos (wt) we get
a = 6 m
w = 8.1 rad/s
a)
w = sqrt (k/m)
8.1 = sqrt (54 / m)
m = 0.823 kg
b)
Time period T = 2*pi/w
= 2*3.14/8.1
= 0.775 s
c)
Max speed = aw = 6*8.1 = 48.6 m/s
d)
Max potential energy = 1/2*ka^2 = 1/2*54*6^2 = 972 J
e)
Total energy = Max. potential energy = 972 J
2.
Potential energy at highest point = mg*0.25L
KE at lowest point = 1/2*mv^2
Energy conservation: 1/2*mv^2 = mg*0.25L
v = sqrt (gL/2)
3.
Max. potential energy = 1/2*ka^2 = 1/2*274*0.13^2 = 2.32 J
4.
2 deg = 2*3.14/180 rad = 0.03488 radians
Torque = k*theta
= 1*10^-8 *0.03488
= 0.03488*10^-8 Nm
Force = Torque / L
= 0.03488*10^-8 / 0.14
= 0.249*10^-8 N
Mass of mosquito = 10^-3 grams = 10^-6 kg
Weight of mosquito = mg = 10^-6 *9.81 = 9.81*10^-6 N
w/F = (9.81*10^-6) / (0.249*10^-8)
= 3939.7
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