when a mass attached to a spring is released from rest 3.0 cm from its equilibrium position, it oscillates with a frequency f. If this mass were instead released from rest 6.0 cm from its equilibrium position, it would oscillate with frequency..
A.2f
B.sqrt 2f
C.f
D.f/2
I can do the math part, I am just unsure of what equation to use to obtain the answer. Thanks
frequency of oscillation does not depend on elongation.
Well, let me "spring" into action and help you out with that! When a mass is attached to a spring, the frequency of its oscillation is determined by the equation:
f = 1 / (2π√(m/k))
where f is the frequency, m is the mass, and k is the spring constant.
Now, let's consider the situation where the mass is initially released from rest 3.0 cm from its equilibrium position. Using the given information, you can calculate the frequency using the equation above.
But, what happens if we change the initial displacement to 6.0 cm? The frequency will actually remain the same! The frequency of oscillation depends on the mass and the spring constant, but not on the initial displacement. So, the answer is C. f.
Remember, when it comes to oscillations, it's all about "bobbin' up and down" without any preference for how you started!
To solve this problem, you can use the equation for the frequency of a mass-spring system:
f = (1 / 2π) * sqrt(k / m)
Where:
- f is the frequency of the oscillation
- k is the spring constant
- m is the mass
The formula shows that the frequency depends on the square root of the ratio of the spring constant to the mass.
In this case, the mass is constant, so we can assume that the spring constant (k) remains the same as well.
Now let's consider the two scenarios:
For the first scenario, when the mass is released from rest 3.0 cm from its equilibrium position, the frequency is f.
For the second scenario, when the mass is released from rest 6.0 cm from its equilibrium position, let's denote the frequency as f'. Since the mass and spring constant are the same, the frequency only depends on the square root of the ratio of the initial displacement (3.0 cm) to the new initial displacement (6.0 cm).
So, mathematically, we can write:
f' = (1 / 2π) * sqrt(k / m) * sqrt(3.0 cm / 6.0 cm)
= (1 / 2π) * sqrt(k / m) * sqrt(1/2)
= (1 / 2π) * sqrt(k / m) * (1 / sqrt(2))
= (1 / sqrt(2π)) * sqrt(k / m)
Simplifying further, we can write:
f' = (1 / sqrt(2)) * f
Therefore, the frequency when the mass is released from rest 6.0 cm from its equilibrium position is (1 / sqrt(2)) times the original frequency. So, the correct answer is option B: sqrt(2f).