A 0.22 kg mass attached to a spring undergoes simple harmonic motion with a period of 0.45 s. what is the force constant of the
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Answer:
1) 341 Hz
Explanation:
When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.
The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.
If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.
The frequency of the n-th harmonic is given by
where
n is the number of the harmonic
is the fundamental frequency
Here we have:
is the frequency of the 3rd harmonic
So the fundamental frequency is
And so, the frequency of the 2nd harmonic is:
The mass of the first block will be three times the mass of the second block.
According to Newton's second law of motion, the force acting on a body is directly proportional to the acceleration as shown;
F is the acting force
m is the mass
a is the acceleration of the body
Given the following parameters
Constant force F = 1N
For the first block with the acceleration of "a"
1 = m₁a
a = m₁/1
m₁ = a .................1
For the second block, acceleration is thrice that of the first. This means;
F = m(3a)
1 = 3ma
..........................2
Divide both equations
From the calculation, we can conclude that the mass of the first block will be three times the mass of the second block.
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