1.A simple pendulum has a length of 2.1 m and is pulled a distance y = 0.24 m to
ID: 2300908 • Letter: 1
Question
1.A simple pendulum has a length of 2.1 m and is pulled a distance y = 0.24 m to one side and then released (see figure below).
(a) What is the speed of the pendulum when it passes through the lowest point on its trajectory?
(b) What is its acceleration in the direction along its circular arc when it passes through the lowest point on its trajectory?
2.Consider a torsional oscillator like the one in the figure below and suppose it has a frequency of 0.076 Hz. If the length of the (massless) rod is increased by a factor of four, what is the new frequency of the oscillator?
Explanation / Answer
1.
length L=2.1m
amplitude A =0.24 m
a)
the speed of the pendulum when it passes through the lowest point on its trajectory is maximum
and is equal to V=wA =sqrt(g/L)*A =sqrt(9.8/2.1)*0.24 =0.518 m/s
b)
acceleration a =0
2.
time peroid of the torsonal pendulum
f =(1/2pi)*sqrt(mgd/I)
I = moment of inertia=m(L/2)^2+ m(L/2)^2 =mL^2/2
if length increased by a factor 4
final lenght L2=4L
that is each mass is at a distance of 2L from the axis of rotation
so new moment of inertia I'=m(2L)^2+m(2L)^2=8ml^2
moment of inertia increases by 16 times
f' =f =(1/2pi)*sqrt(mgd/I')
so frequency decreases by 4 times f'=f/4=0.076/4 =0.019 Hz
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