Damped Oscillations: Mass 300 g, k=1.50 N/m, b in kg/s

Kinetic energy graphs would look like a sine function, with the amplitude decreasing and phase shifting over time. In summary, the weight of 300 g hanging from a spring with a spring constant of 1.50 N/m undergoes vertical oscillation with an initial amplitude that decreases to 1/10 of its original value after 28 seconds. The damping constant associated with this motion is unknown and the graphs of the mechanical and kinetic energies would resemble cosine and sine functions, respectively. The equation of motion is x = A_0e^{-qt}\sin(\omega t + \phi).
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physicsgurl123
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A weight of mass m = 300 g hangs vertically from a spring that has a spring constant k = 1.50 N/m. The mass is set into vertical oscillation and after 28 s you find that the amplitude of the oscillation is 1/10 that of the initial amplitude. What is the damping constant b associated with the motion (in kg/s)? Also what would the graphs of the mechanical and kinetic energies look like?
 
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physicsgurl123 said:
A weight of mass m = 300 g hangs vertically from a spring that has a spring constant k = 1.50 N/m. The mass is set into vertical oscillation and after 28 s you find that the amplitude of the oscillation is 1/10 that of the initial amplitude. What is the damping constant b associated with the motion (in kg/s)? Also what would the graphs of the mechanical and kinetic energies look like?
What is the equation of motion here? Hint: The general form of solution is:

[tex]x = A_0e^{-qt}\sin(\omega t + \phi)[/tex]AM
 
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Based on the given information, we can use the equation for damped oscillations to determine the damping constant b:

mω = b/2m

Where m is the mass of the weight (300 g) and ω is the angular frequency, which can be calculated using the formula:

ω = √(k/m)

Substituting the values, we get:

ω = √(1.50 N/m / 0.3 kg) = √5 rad/s

Now, using the given time and amplitude information, we can calculate the new angular frequency ω' using the formula:

ω' = ω / (1/10)

ω' = 10ω = 50 rad/s

Substituting this value into the equation for damping constant, we get:

b = 2mω' = 2 x 0.3 kg x 50 rad/s = 30 kg/s

Therefore, the damping constant b associated with the motion is 30 kg/s.

As for the graphs of mechanical and kinetic energies, the mechanical energy graph would show a decrease in amplitude over time, as the energy is being dissipated due to damping. The kinetic energy graph would also show a decrease in amplitude, but at a faster rate compared to the mechanical energy graph, as the kinetic energy is directly proportional to the square of the amplitude. Eventually, both graphs would reach a steady state where the energy remains constant, indicating that the oscillations have stopped due to the damping force balancing out the restoring force of the spring.
 

FAQ: Damped Oscillations: Mass 300 g, k=1.50 N/m, b in kg/s

What is a damped oscillation?

A damped oscillation is a type of motion where an object (in this case, a mass of 300 g) oscillates back and forth due to a restoring force (in this case, a spring with a stiffness of 1.50 N/m) but is also subject to a damping force (in this case, a damping coefficient of b in kg/s). This results in the amplitude of the oscillations decreasing over time until the object reaches a state of equilibrium.

How is the damping coefficient (b) related to the rate of decay of the oscillations?

The damping coefficient (b) directly affects the rate of decay of the oscillations. The larger the value of b, the faster the oscillations will decay. This is because a higher damping coefficient means there is more resistance to the motion of the object, resulting in a quicker loss of energy and a shorter period of oscillation.

What is the formula for calculating the period of a damped oscillation?

The formula for calculating the period of a damped oscillation is T = 2π/ω, where T is the period and ω is the angular frequency. The angular frequency is calculated by taking the square root of the stiffness (k) divided by the mass (m) minus half of the damping coefficient (b/2m).

How does changing the mass affect the damped oscillations?

Changing the mass will affect the damped oscillations in a few ways. Firstly, a larger mass will result in a longer period of oscillation. Secondly, a larger mass will also result in a lower angular frequency, meaning the oscillations will occur at a slower rate. Finally, a larger mass will also result in a slower rate of decay of the oscillations.

Can a damped oscillation system ever reach a state of permanent oscillation?

No, a damped oscillation system will never reach a state of permanent oscillation. This is because the damping force will always act to decrease the amplitude of the oscillations, eventually bringing the object to a state of equilibrium where there is no more oscillation. However, the time it takes to reach this state of equilibrium will vary depending on the values of the mass, stiffness, and damping coefficient.

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