A 75 g bungee cord has an equilibrium length of 1.20 m. The cord is stretched to a length of

1.80 m, then vibrated at 20 Hz. This produces a standing wave with two antinodes. What is the
spring constant of the bungee cord?

A standing wave with two antinodes has a wavelength for traveling waves equal to the cord length, 1.80 m.

The wave speed V is the product of wavelength and frequency, or 36 m/s

From that wave speed, derive the Tension required to stretch the cord 0.60 m. From that you can get the spring constant.

Note that the lineal density of the stretched cord is
d = 0.075 kg/1.80 m
= 4.167*10^-2 kg/m.

You will need that to relate the wave speed to the tension.

V = sqrt(T/d)

T = d*V^2 = 36^2*4.167*10^-2 = 54 N

Spring constant = 54N/0.60 m = 90 N/m

Well, choosing a spring constant for a bungee cord is a bit like choosing a life partner - it's all about finding the right balance! But let me clown around and calculate this for you.

Now, the frequency of the wave is given as 20 Hz, which means the cord is oscillating back and forth 20 times per second. Since there are two antinodes, that means the wavelength of the wave is twice the length of the cord, which is 2.40 m.

Now, using the equation v = fλ, where v is the velocity of the wave, f is the frequency, and λ is the wavelength, we can determine the velocity. So, v = 20 Hz * 2.40 m = 48 m/s.

The velocity of a wave on a string is also given by the equation v = sqrt(T/μ), where T is the tension and μ is the mass per unit length. In this case, we are given the mass of the cord (75 g) and the equilibrium length (1.20 m).

First, let's convert the mass to kilograms: 75 g = 0.075 kg. Now, let's calculate the mass per unit length: μ = 0.075 kg / 1.20 m = 0.0625 kg/m.

Now, we can rearrange the equation to solve for the tension: T = v^2 * μ. Plugging in the values, we get T = (48 m/s)^2 * 0.0625 kg/m = 144 N.

Finally, the spring constant, k, is given by the equation k = T/L, where L is the equilibrium length. Plugging in the values, we have k = 144 N / 1.20 m = 120 N/m.

So, the spring constant of the bungee cord is 120 N/m. Just remember, when it comes to bungee jumping, always take the spring constant seriously!

To find the spring constant of the bungee cord, we can apply Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

Hooke's Law formula: F = k * x

Where:
F is the force exerted by the spring,
k is the spring constant,
x is the displacement from the equilibrium position.

Given:
Mass of the bungee cord (m) = 75 g = 0.075 kg
Equilibrium length of the bungee cord (L0) = 1.20 m
Length of the bungee cord when stretched (L) = 1.80 m
Frequency of vibration (f) = 20 Hz
Number of antinodes (n) = 2

Step 1: Calculate the wavelength (λ) of the standing wave.

Wavelength formula: λ = 2L/n

Substituting the given values:
λ = 2 * 1.80 m / 2 = 1.80 m

Step 2: Calculate the velocity (v) of the wave.

Velocity formula: v = λ * f

Substituting the given values:
v = 1.80 m * 20 Hz = 36 m/s

Step 3: Calculate the tension (T) in the bungee cord.

Tension formula: T = m * g

where g is the acceleration due to gravity,

Substituting the given values:
T = 0.075 kg * 9.8 m/s^2 = 0.735 N

Step 4: Calculate the angular frequency (ω) of the wave.

Angular frequency formula: ω = 2πf

Substituting the given value:
ω = 2 * π * 20 Hz = 40π rad/s

Step 5: Calculate the spring constant (k) using the formula:

k = mω^2

Substituting the given value:
k = 0.075 kg * (40π rad/s)^2 ≈ 94.247 N/m

Therefore, the spring constant of the bungee cord is approximately 94.247 N/m.

To find the spring constant of the bungee cord, we can use the formula for the frequency of a standing wave on a string:

f = (1/2L) * sqrt(T/µ)

Where:
- f is the frequency of the wave
- L is the length of the string
- T is the tension in the string
- µ is the linear mass density of the string

In this case, the string is the bungee cord. We are given that the equilibrium length of the cord is 1.20 m and it is stretched to a length of 1.80 m. This means that the amplitude of vibration is (1.80 m - 1.20 m)/2 = 0.30 m.

Since the standing wave has two antinodes, the length of the string from one antinode to the other is one-half of the wavelength. Therefore, the length of the cord between two antinodes is L = 0.30 m.

We are also given the frequency of vibration, which is 20 Hz.

Now, we can substitute the known values into the formula and solve for the tension T:

20 Hz = (1/2 * 0.30 m) * sqrt(T/µ)

Rearranging the formula, we get:

sqrt(T/µ) = (20 Hz) * (2 * 0.30 m)

Simplifying:

sqrt(T/µ) = 12

Now, we square both sides to eliminate the square root:

T/µ = 144

Since we want to find the spring constant of the bungee cord, which is represented by k, we can relate the tension to the spring constant using the equation T = k * Δx. Where Δx is the displacement from the equilibrium position.

The mass of the cord is given as 75 g, which is equivalent to 0.075 kg. Since the tension T is equal to the weight of the cord at equilibrium, we can calculate:

T = mg = (0.075 kg) * (9.8 m/s^2) = 0.735 N

Now, substituting T = 0.735 N into the tension equation:

0.735 N = k * (1.80 m - 1.20 m)

0.735 N = k * 0.60 m

Solving for k:

k = 0.735 N / 0.60 m

k ≈ 1.22 N/m

Therefore, the spring constant of the bungee cord is approximately 1.22 N/m.