The length of a simple pendulum is 0.70 m and the mass of the particle ( the "bob") at the end of the cable is 0.33 kg. The pendulum is pulled away from ts equilibrium position by an angle of 7.2 degrees and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings abck and forth. (c) what is the bob's speed as it passes through the lowest point of the swing?

Question
Asked Nov 5, 2019

The length of a simple pendulum is 0.70 m and the mass of the particle ( the "bob") at the end of the cable is 0.33 kg. The pendulum is pulled away from ts equilibrium position by an angle of 7.2 degrees and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings abck and forth. (c) what is the bob's speed as it passes through the lowest point of the swing?

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Step 1

Given,

Length of pendulum, 0.70 m
Mass of bob,m0.33 Kg
Angle displaced,0=7.2°
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Length of pendulum, 0.70 m Mass of bob,m0.33 Kg Angle displaced,0=7.2°

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Step 2

Part (a):

Angular frequency (ω) of simple pendulum can be given as,

1 g
2T
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1 g 2T

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Step 3

By substituting the ...

1
f
2л V1
9.80
f
2л \0.70
1
1
f =
(0.60)=0.095 Hz
2л
Ф-2л/
о-2я(0095)
о -0.595 гad/sec
ఉం
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1 f 2л V1 9.80 f 2л \0.70 1 1 f = (0.60)=0.095 Hz 2л Ф-2л/ о-2я(0095) о -0.595 гad/sec ఉం

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