A simple pendulum with mass m = 1.8 kg and length L = 2.66 m hangs from the ceil
ID: 1506487 • Letter: A
Question
A simple pendulum with mass m = 1.8 kg and length L = 2.66 m hangs from the ceiling. It is pulled back to an small angle of = 8.6° from the vertical and released at t = 0.
1)
What is the period of oscillation?
s
2)
What is the magnitude of the force on the pendulum bob perpendicular to the string at t=0?
N
3)
What is the maximum speed of the pendulum?
m/s
4)
What is the angular displacement at t = 3.63 s? (give the answer as a negative angle if the angle is to the left of the vertical)
°
5)
What is the magnitude of the tangential acceleration as the pendulum passes through the equilibrium position?
m/s2
6)
What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position?
m/s2
7)
Which of the following would change the frequency of oscillation of this simple pendulum?
increasing the mass
decreasing the initial angular displacement
increasing the length
hanging the pendulum in an elevator accelerating downward
Explanation / Answer
given that
m = 1.8 kg
L = 2.66 m
theta = 8.6 degree
(a)
we know that
period T = 2*pi*sqrt(L/g)
T = 2*3.14*sqrt(2.66/9.8) = 3.27 s
(b)
magnitude of the force on the pendulum
F = m*g*sin(theta)
F = 1.8*9.8*sin(8.6) = 2.63 N
(c)
from energy conservation
PE = KE
m*g*h = 1/2 * m*v^2
v = sqrt(2*g*h)
h = L-L*cos(theta) = 2.66-(2.66*cos8.6)
h = 0.03 m
spees v = sqrt(2*9.8*0.03) = 0.76 m/s
(d)
given
t = 3.63 s
angular displacement = initial angle * cos (sqrt(g/L)*t)
angular displacement = 8.6*cos(sqrt(9.8/2.66)*3.5) = 6.70 deg
(e)
as the only acceleration acting on pendulum gravitational acceleration, which acts vertically
so magnitude of the tangential acceleration = 0
(f)
magnitude of the radial acceleration = gravitational acceleration = g
ar = 9.81 m/s^2
(g)
we know that
frequency of oscillation f = k*sqrt(g/L)
it will only change by
(c) increasing the length
(d) hanging the pendulum in an elevator accelerating downward
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