As in the figure below, a simple harmonic oscillator is attached to a rope of linear mass density 5.4 ✕ 10^−2 kg/m, creating a standing transverse wave. There is a 3.2-kg block hanging from the other end of the rope over a pulley. The oscillator has an angular frequency of 44.3 rad/s and an amplitude of 256.0 cm.
(a) What is the distance between adjacent nodes?
(b) If the angular frequency of the oscillator doubles, what happens to the distance between adjacent nodes? (Enter the new distance.)
(c) If the mass of the block is doubled instead, what happens to the distance between adjacent nodes? (Enter the new distance.)
(d) If the amplitude of the oscillator is doubled, what happens to the distance between adjacent nodes? (Enter the change in distance.)
To find the answers to these questions, we need to understand the properties of standing waves on a string and how they depend on various parameters.
Let's start by explaining some key concepts.
1. Nodes: These are points on the string where there is no motion. The amplitude of the wave is minimum at these points.
2. Antinodes: These are points on the string with maximum displacement and maximum amplitude.
3. Wavelength (λ): It is the distance between two adjacent nodes (or antinodes).
Now, let's solve the given questions:
(a) What is the distance between adjacent nodes?
The distance between adjacent nodes is equal to half the wavelength (λ/2) of the standing wave.
To calculate the wavelength, we can use the formula:
λ = 2π / k
where k is the wave number, given by:
k = ω / v
Here, ω represents the angular frequency and v represents the velocity of the wave on the string.
Given information:
Angular frequency (ω) = 44.3 rad/s
Linear mass density (μ) = 5.4 * 10^(-2) kg/m
The velocity (v) of the wave on the string can be determined using the formula:
v = (T / μ)^(1/2)
where T is the tension in the string.
We don't have the value of tension, but we need to consider the gravitational force acting on the hanging block.
The gravitational force (F_g) is given by:
F_g = m * g
where m is the mass of the block and g is the acceleration due to gravity.
Given information:
Mass of the block (m) = 3.2 kg
Acceleration due to gravity (g) = 9.8 m/s^2
The tension (T) in the string is equal to the gravitational force acting on the block.
T = F_g = m * g
Now we can calculate the velocity (v) using the tension (T) and linear mass density (μ).
Calculate T:
T = m * g = 3.2 kg * 9.8 m/s^2
Calculate v:
v = (T / μ)^(1/2)
Now, substitute the values of T and μ into the formula to find v.
Once we have v, we can calculate the wavelength (λ) using the formula mentioned earlier:
λ = 2π / k
Finally, divide the obtained wavelength by 2 to find the distance between adjacent nodes.
(b) If the angular frequency of the oscillator doubles, what happens to the distance between adjacent nodes?
To determine the new distance between adjacent nodes, we need to keep in mind that the velocity (v) of the wave on the string remains the same since the tension and linear mass density is unchanged. Therefore, the wavelength (λ) will be halved and the distance between adjacent nodes will also be halved.
(c) If the mass of the block is doubled, what happens to the distance between adjacent nodes?
Doubling the mass of the block will increase the tension (T) in the string. As a result, the velocity (v) of the wave on the string will increase. Since λ = 2π / k, and k depends on v, increasing v will decrease the wavelength (λ), and consequently, the distance between adjacent nodes will also decrease.
(d) If the amplitude of the oscillator is doubled, what happens to the distance between adjacent nodes?
The amplitude of the oscillator does not affect the distance between adjacent nodes. The wavelength and distance between the nodes remain the same.
To find the distance between adjacent nodes in a standing transverse wave, we can use the formula:
Distance between adjacent nodes = λ/2
where λ is the wavelength of the wave.
(a) To find the distance between adjacent nodes, we need to find the wavelength first.
We know that the linear mass density of the rope is 5.4 ✕ 10^−2 kg/m, so the mass per unit length of the rope (μ) is 5.4 ✕ 10^−2 kg/m.
The angular frequency of the oscillator is 44.3 rad/s, so the frequency of the wave (f) is given by:
f = ω/2π
f = 44.3/2π ≈ 7.04 Hz
The speed of the wave (v) can be calculated using the formula:
v = fλ
where v is the speed of the wave and λ is the wavelength.
Since the rope is under tension due to the 3.2-kg block hanging from the other end over a pulley, the speed of the wave is given by:
v = sqrt(T/μ)
where T is the tension in the rope.
The tension in the rope can be calculated using the formula:
T = mg
where m is the mass of the block and g is the acceleration due to gravity.
T = (3.2 kg) * (9.8 m/s^2) ≈ 31.36 N
So, the speed of the wave is:
v = sqrt(31.36 N / (5.4 ✕ 10^−2 kg/m))
Now, we can find the wavelength:
v = fλ
λ = v/f
Plug in the values:
λ = (sqrt(31.36 N / (5.4 ✕ 10^−2 kg/m))) / (7.04 Hz)
Calculate the value of λ.
(b) If the angular frequency of the oscillator doubles, the frequency of the wave (f) will double as well. Using the equation:
v = fλ
We can see that v will remain the same because the tension in the rope and the linear mass density of the rope have not changed.
So, the wavelength (λ) remains the same for the doubled angular frequency.
(c) If the mass of the block is doubled, the tension in the rope (T) will double as well.
Using the equation:
v = sqrt(T/μ)
We can see that v will increase because the tension in the rope has increased, but the linear mass density of the rope remains the same.
So, the wavelength (λ) will decrease because v increases and f remains the same.
(d) If the amplitude of the oscillator is doubled, it will not affect the distance between adjacent nodes. The amplitude only affects the maximum displacement of the wave, not the distance between adjacent nodes.