A simple harmonic oscillator oscillates with frequency f when its amplitude is
1 answer:
If the amplitude is doubled, the frequency is still f.
In fact, the frequency of a simple harmonic oscillator does not depend on the amplitude. The frequency of the oscillator in fact is:
where
k is the spring constant
m is the mass attached to the spring
We can see that the frequency does not depend on the amplitude A, but only on the properties of the oscillator, so if the amplitude is doubled, the frequency does not change.
In fact, the frequency of a simple harmonic oscillator does not depend on the amplitude. The frequency of the oscillator in fact is:
where
k is the spring constant
m is the mass attached to the spring
We can see that the frequency does not depend on the amplitude A, but only on the properties of the oscillator, so if the amplitude is doubled, the frequency does not change.
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Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = = -
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object