Simple harmonics - sound waves

Your name] In summary, using the given information and the equation v = fλ, we can calculate the wavelength of a 2.40 kHz sound wave in aluminum as 2.125 m. When the sound wave passes into air, the frequency remains the same and the wavelength becomes 2.125 m as well.
  • #1
Foxhound101
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0

Homework Statement


A 2.40 kHz sound wave is transmitted through an aluminum rod.

A) What is its wavelength in the aluminum?
B What is the sound wave's frequency when it passes into the air?
C) What is its wavelength in air?

The speed of sound in aluminum is 5100 m/s according to the book.

Homework Equations



Velocity=frequency*lambda


The Attempt at a Solution



So...This is what I was doing, which apparently is wrong.

Velocity=frequency*lambda

frequency=2.4 * 10^3 Hz
Velocity = 5100 meters/sec

5100m/s=(2.4 * 10^3)hz *lambda

lambda=2.125m

But apparently that isn't the answer for part A...So I must have done something wrong somewhere...

So...thank you in advance for any help.
 
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  • #2

Thank you for your question. I would be happy to assist you in finding the correct solution.

First, let's review the given information and equations. The speed of sound in a material can be calculated using the equation v = fλ, where v is the speed of sound, f is the frequency, and λ is the wavelength.

For part A, we are given the frequency and the speed of sound in aluminum, and we need to find the wavelength. Therefore, we can rearrange the equation to solve for λ, which gives us λ = v/f.

Substituting the given values, we get λ = (5100 m/s) / (2.40 kHz) = 2.125 m. This is the correct answer for the wavelength in aluminum.

For part B, the sound wave passes from aluminum into air. The speed of sound in air is approximately 343 m/s. Using the same equation, we can calculate the frequency in air as f = v/λ = (343 m/s) / (2.125 m) = 161.41 Hz. This is the answer for part B.

Finally, for part C, we need to find the wavelength in air. Again, using the same equation, we get λ = v/f = (343 m/s) / (161.41 Hz) = 2.125 m. This is the same as the wavelength in aluminum, as the frequency does not change when the sound wave passes into a different medium.

I hope this helps clarify the solution for you. If you have any further questions, please do not hesitate to ask.
 
  • #3


Your approach is correct, but you have used the wrong velocity value. The given velocity of 5100 m/s is for sound waves in aluminum, not for air. The velocity of sound in air is approximately 343 m/s. So, your calculation should be:

Velocity = frequency * lambda
343 m/s = (2.4 * 10^3) Hz * lambda
lambda = 0.143 m

Therefore, the wavelength of the sound wave in aluminum is 0.143 meters and in air is 2.125 meters.
 

Related to Simple harmonics - sound waves

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around a central equilibrium point. It is characterized by a sinusoidal pattern and is governed by Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement of the object from its equilibrium position.

2. How is simple harmonic motion related to sound waves?

Sound waves are a type of mechanical wave that can be described as simple harmonic motion. The compressions and rarefactions of the sound wave follow a sinusoidal pattern, just like an object undergoing simple harmonic motion. This is because sound waves are produced by the vibration of an object, which creates a periodic back and forth motion.

3. What is the relationship between frequency and pitch in sound waves?

Frequency is the number of cycles per second in a sound wave, and pitch is how high or low we perceive a sound to be. The higher the frequency of a sound wave, the higher the pitch will be. This is because our brains interpret higher frequencies as higher pitches.

4. How do amplitude and energy relate in sound waves?

Amplitude is the measure of the displacement of particles in a sound wave, while energy is the ability to do work. In sound waves, the amplitude determines the loudness of the sound, and the energy of the wave is directly proportional to the square of the amplitude. This means that as the amplitude increases, the energy and loudness of the sound also increase.

5. Can simple harmonic motion be observed in everyday life?

Yes, simple harmonic motion can be observed in many everyday objects and phenomena. Examples include the motion of a pendulum, the swinging of a child on a swing, and the vibrations of a guitar string. It is also present in natural phenomena such as ocean waves and the movement of celestial bodies.

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