A 0.363 g wire is stretched between two points 80.3 cm apart. The tension in the wire is 573 N.
Find the frequency of first harmonic. Answer in units of Hz
λ=v/f => f = v/ λ,
v = sqrt(TL/m ),
λ=2L,
f = sqrt(TL/m )/ 2L=
=sqrt(573•0.803/0.363•10⁻³)/2•0.803=
=701 Hz
Well, isn't this a stretchy situation? Let's have some fun with this physics problem!
To find the frequency of the first harmonic, we need to remember a little bit about waves on a string. The fundamental frequency of a stretched wire can be found using the formula:
f = (1/2L) * sqrt(T/m)
where f is the frequency, L is the length of the wire, T is the tension in the wire, and m is the mass per unit length of the wire.
In this case, we have L = 80.3 cm and T = 573 N. But where's m, you ask? Well, don't worry, I've got it covered!
To find m, we need to know the mass of the wire. Since we're given the mass of the wire per unit length, all we need to do is multiply it by the length of the wire. So let's get multiplying!
m = (0.363 g / 100 cm) * 80.3 cm
Calculating that, we get m ≈ 0.291 g/cm.
Now, let's substitute the values into the frequency formula:
f = (1/2 * 80.3 cm) * sqrt(573 N / (0.291 g/cm))
After some calculations, we find that the frequency of the first harmonic is approximately 159 Hz.
So there you have it - the first harmonic is singing at about 159 Hz! Remember to stretch, but not too much, or you might snap your funny bone!
To find the frequency of the first harmonic, we can use the formula:
f = (1/2L) * √(T/μ)
Where:
f is the frequency of the harmonic,
L is the length of the wire,
T is the tension in the wire,
μ is the linear density of the wire.
First, we need to find the linear density of the wire, μ. The linear density (μ) of a wire is defined as the mass per unit length of the wire.
Given:
Mass of the wire (m) = 0.363 g = 0.363 * 10^(-3) kg (converting g to kg)
Length of the wire (L) = 80.3 cm = 80.3 * 10^(-2) m (converting cm to m)
Tension in the wire (T) = 573 N
Let's calculate the linear density:
μ = m / L
μ = (0.363 * 10^(-3)) kg / (80.3 * 10^(-2)) m
Now, let's substitute the values into the formula to find the frequency:
f = (1 / 2L) * √(T/μ)
f = (1 / 2 * (80.3 * 10^(-2)) m) * √(573 N / [(0.363 * 10^(-3)) kg / (80.3 * 10^(-2)) m])
Simplifying further:
f = (1 / (2 * 80.3 * 10^(-2))) * √(573 / (0.363 * 10^(-3)))
Now, calculating the value:
f ≈ 166.8 Hz (rounded to one decimal place)
Therefore, the frequency of the first harmonic is approximately 166.8 Hz.
To find the frequency of the first harmonic, we need to use the formula:
f = (1/2L) * √(T/u)
Where:
f is the frequency of the harmonic
L is the length of the wire
T is the tension in the wire
u is the linear density of the wire (mass/length)
First, we need to calculate the linear density (u) of the wire:
u = mass / length
Given:
mass = 0.363 g = 0.000363 kg (since 1 g = 0.001 kg)
length = 80.3 cm = 0.803 m (since 1 m = 100 cm)
So, u = 0.000363 kg / 0.803 m = 0.000451 kg/m
Next, we can substitute the values into the formula to calculate the frequency:
f = (1/2 * 0.803 m) * √(573 N / 0.000451 kg/m)
Now, we can simplify and calculate:
f = 0.5 * 0.803 * √(573 / 0.000451) Hz
f = 0.4015 * √(1,269,401.33) Hz
Using a calculator or software, we find:
f ≈ 48.30 Hz
Therefore, the frequency of the first harmonic is approximately 48.30 Hz.
A 0.373 g wire is stretched between two points
89.2 cm apart. The tension in the wire is
533 N.