Sound Waves and frequency

In summary, the conversation discusses a block with a speaker attached to a spring with a spring constant of 20.0 N/m. The total mass of the block and speaker is 5.00 kg and it has an amplitude of 0.500 m. It is emitting sound waves with a frequency of 440 Hz. By using the equation 1/2 kx^2 = ma and finding the speed of the speaker, the range of frequencies heard by a person to the right of the speaker is determined to be 439 Hz < f < 441 Hz. The conversation also mentions using the equation v1^2 = u^2 + 2as to find the value of v1 and calculating the maximum and minimum frequencies
  • #1
frozen7
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7. A block with a speaker bolted to it is connected to a spring having spring constant k = 20.0 N/m as in Figure 3. The total mass of the block and speaker is 5.00 kg, and the amplitude of this unit’s motion is 0.500 m. If the speaker emits sound waves of frequency 440 Hz, determine the range in frequencies heard by the person to the right of the speaker.
Ans: [ 439 Hz < f < 441 Hz ]


Can anyone help me to solve this? Give me some clues please.
 
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  • #2
I do it in this way:

1/2 k x^2 = ma
a = 0.5
And I found the value of v1 (speed of speaker) = 0.707 by using v1^2 = u^2 =2as

Then I find the max frequency by f`=f ( v / ( v-v1))
min frequescy = f`=f ( v / ( v+v1))


Am I doing correctly?
 
  • #3


Sure, here are some clues to help you solve this problem:

1. Start by understanding the concept of sound waves and frequency. Sound waves are a type of mechanical wave that travels through a medium, such as air, and can be detected by our ears. Frequency is the number of cycles or vibrations of a wave that occur in one second. It is measured in Hertz (Hz).

2. Next, look at the given information in the problem. The speaker is attached to a spring with a spring constant of 20.0 N/m, and the total mass of the block and speaker is 5.00 kg. The amplitude of the motion is also given as 0.500 m.

3. Use the formula for frequency, f = 1/T, where T is the period of the wave. The period is the time it takes for one complete cycle of the wave. In this case, the wave is being produced by the speaker, so the period will be the same as the time it takes for the speaker to complete one full vibration.

4. Calculate the period using the formula T = 2π√(m/k), where m is the mass and k is the spring constant. In this problem, the mass is 5.00 kg and k is 20.0 N/m. Plug these values into the formula and solve for T.

5. Once you have calculated the period, use the formula for frequency to find the range of frequencies heard by the person to the right of the speaker. The formula is f = 1/T, where T is the period you calculated in step 4. This will give you the frequency of the sound wave produced by the speaker.

6. Keep in mind that the human ear can typically hear frequencies between 20 Hz and 20,000 Hz. So, if the calculated frequency falls within this range, then the range in frequencies heard by the person to the right of the speaker will be [calculated frequency - 1 Hz < f < calculated frequency + 1 Hz]. If the calculated frequency is outside of this range, then the range of frequencies heard will be [20 Hz < f < 20,000 Hz].

I hope these clues help you solve the problem. Good luck!
 

1. What is a sound wave?

A sound wave is a type of mechanical wave that is created by vibrations and travels through a medium, such as air, water, or solids. It is characterized by its frequency, amplitude, and wavelength.

2. How is sound frequency measured?

Sound frequency is measured in Hertz (Hz), which represents the number of cycles or vibrations per second. The human hearing range is typically between 20 Hz to 20,000 Hz.

3. What affects the frequency of a sound wave?

The frequency of a sound wave is affected by the source of the sound, such as the frequency of a vibrating object, and the medium through which it travels. The density and temperature of the medium can also affect the frequency of a sound wave.

4. How does sound frequency impact pitch?

Sound frequency is directly related to pitch. Higher frequency sound waves have a higher pitch, while lower frequency sound waves have a lower pitch. This is why musical notes have different frequencies and produce different pitches.

5. Can sound waves travel through a vacuum?

No, sound waves cannot travel through a vacuum because they require a medium to propagate. Therefore, sound cannot travel in outer space where there is no air or other medium.

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