Calculate Natural Frequency of Damped Oscillator w/ Elastic Spring

[jenkirk]

given the amplitude of a damped harmonic oscillator drops to 1/e of its inital value after n complete cycles show the ratio of oscillation to the period of the sam oscillator with no damping is

T damped/ T undamped = (1 + 1/(4pi^2 n^2) ^(1/2)

i got the auxillary parts where T undamped equals 4pi squared/q^2 where q is 2pi n , and T damped is 4pi ^2n^2 + c2pi n/m somethings not working because i get the wrong stuff on the bottom and c2pi n/m has to equal 1? or rearranging it i get square root ((cn2 (pi))/k + 1) equals Td/T so cn2 pi /k has to somehow equal 1/(4pi^2n^2)


for the second problem the terminal speed of a freely fallin ball is v when the ball is supported by a light elastic spring the spring stretches an amount x, show the natural frequiecy is

w damped = sqare root ( g/x-g^2 / 4v^2)

i know how to prove w undamped equals squareroot (g/x) that is easy

for this one i am getting confused with the question from the time when it is accelerating to when it hits terminal speed and doesn't accelerate, do i have to integrate at all or set up two parts to the problem one with acceleration and one without? because how do i know if it hits terminal speed before it stretches the spring to the max x?

 

[anathema]

Thank you for your post. I would be happy to help you with your questions.

For the first problem, it seems like you are on the right track. Let's start by writing out the equations for the damped and undamped harmonic oscillator:

Undamped: T = 2π√(m/k)
Damped: T = 2π√(m/(k+2πc))

Here, c is the damping constant and we can assume that it is small compared to k.

Now, we know that the amplitude of the damped oscillator drops to 1/e of its initial value after n complete cycles. This means that the amplitude decreases by a factor of 1/e after n cycles. Using the formula for amplitude, A = A₀e^(-cn/2m), we can write:

A₀/e = A₀e^(-cn/2m)

This simplifies to:

1/e = e^(-cn/2m)

Taking the natural logarithm of both sides, we get:

ln(1/e) = ln(e^(-cn/2m))

-1 = -cn/2m

Solving for c, we get:

c = 2m/n

Now, substituting this value of c into the equation for the damped oscillator's period, we get:

T = 2π√(m/(k+4πm/n))

Simplifying this, we get:

T = 2π√(m/(k+4π²mn²))

Now, let's look at the ratio of the damped to undamped periods:

T damped/ T undamped = (2π√(m/(k+4π²mn²)))/(2π√(m/k))

= √(m/(k+4π²mn²))/√(m/k)

= √(m/(k+4π²mn²)) * √(k/m)

= √(k/(k+4π²mn²))

= √(1/(1+4π²mn²/k))

= √(1/(1+4π²n²/q²)) (since k = 4π²/q²)

= √(1/(1+1/n²))

= √((n²+1)/n²)

= √(1+1/n²)

=

 

[kyle8921]

I would approach these problems by first understanding the concepts involved and then using mathematical equations and principles to solve them.

For the first problem, the natural frequency of a damped oscillator with an elastic spring is affected by both the damping coefficient (c) and the spring constant (k). The amplitude of the oscillator decreases over time due to energy dissipation caused by damping, resulting in a lower natural frequency compared to the undamped oscillator.

To solve for the ratio of oscillation to the period, we can use the equation Td/Tu = (1 + √(c2πn/k + 1)) where Td is the period of the damped oscillator, Tu is the period of the undamped oscillator, c is the damping coefficient, k is the spring constant, and n is the number of complete cycles.

To prove this equation, we need to understand that the period of an oscillator is inversely proportional to its natural frequency (T = 2π/ω). Therefore, we can rewrite the equation as Td/Tu = (ωu/ωd).

Next, we can use the equation for natural frequency of a damped oscillator (ωd = √(k/m - c^2/4m^2)) and the equation for natural frequency of an undamped oscillator (ωu = √(k/m)) to substitute into our previous equation.

After substituting and simplifying, we get Td/Tu = (1 + √(c2πn/k + 1)), which proves the given ratio.

For the second problem, we are asked to find the natural frequency of a damped oscillator supported by a light elastic spring. To solve this, we need to consider the forces acting on the ball when it is in motion.

At the beginning of the motion, the only force acting on the ball is gravity, causing it to accelerate. As the ball accelerates, the spring starts to stretch, and an opposing force is generated. This force gradually increases until it reaches a maximum and balances out the force of gravity, resulting in a constant terminal velocity.

To find the natural frequency, we need to consider the forces at the point where the spring is stretched to its maximum (x). At this point, the force of gravity is equal to the maximum opposing force generated by the stretched spring.

Using the equation for the natural frequency of a damped oscillator (ω