What are the lowest two frequencies of the sound?

  • Thread starter Thread starter Yoda13
  • Start date Start date
  • Tags Tags
    Frequencies Sound
AI Thread Summary
The discussion focuses on calculating the lowest two frequencies of sound emitted by a loudspeaker based on the speed of sound (340 m/s) and the distances to two listeners at specified coordinates. By determining the wavelengths corresponding to each listener's distance—40m for one and 30m for the other—the frequencies can be calculated using the formula frequency = speed of sound / wavelength. This results in frequencies of 8.5 Hz and 11.3 Hz for the two listeners. The problem illustrates the relationship between sound waves, distance, and frequency. Understanding these concepts is essential for solving similar sound wave problems.
Yoda13
Messages
2
Reaction score
0
I am having a problem finding where to start. The problem is this: A loudspeaker at the orgin emits sound waves on a day where the speed of sound is 340 m/s. A crest of the wave simultaneously passes listeners at the (x,y) coordinates (40m,0m) and (0m,30m). What are the lowest two frequencies of the sound? If the question asks for do different frequecies then there must be two different waves, right? I am having trouble thinking how to relate anything to finding the frequecy. Thanks for your help.
 
Physics news on Phys.org
Yoda13 said:
I am having a problem finding where to start. The problem is this: A loudspeaker at the orgin emits sound waves on a day where the speed of sound is 340 m/s. A crest of the wave simultaneously passes listeners at the (x,y) coordinates (40m,0m) and (0m,30m). What are the lowest two frequencies of the sound? If the question asks for do different frequecies then there must be two different waves, right? I am having trouble thinking how to relate anything to finding the frequecy. Thanks for your help.
Well, you know the speed, you have enough info to find the wavelength, and all you need to find is the frequency. To find the wavelength, place the 2 coordinates on a graph. (You know that waves travel in sinusoidal fashion, right?) Now, there are 2 possibilities for attaching the 2 points together. They could represent a quarter of a wavelenth, or half a wavelength. Now that I gave you a hint, you should be able to solve the problem.
 


To find the lowest two frequencies of the sound, we need to first understand the concept of frequency. Frequency is defined as the number of cycles or vibrations per second, and is measured in Hertz (Hz). In the context of sound, frequency is related to the pitch of the sound, with higher frequencies producing higher pitched sounds.

In this problem, we are given the speed of sound (340 m/s) and the coordinates of the listeners (40m,0m) and (0m,30m). This means that the distance between the loudspeaker and the first listener is 40m, and the distance between the loudspeaker and the second listener is 30m.

To find the frequency of the sound, we can use the formula: frequency = speed of sound / wavelength. The wavelength is the distance between two consecutive crests or troughs of a wave. In this case, we have two listeners at different distances from the loudspeaker, so we will have two different wavelengths.

For the first listener at (40m,0m), the distance between the loudspeaker and the listener is 40m. This means that the wavelength is also 40m. Plugging in the given values into the formula, we get:

Frequency = 340 m/s / 40m = 8.5 Hz

Similarly, for the second listener at (0m,30m), the distance between the loudspeaker and the listener is 30m. This means that the wavelength is 30m. Plugging in the given values into the formula, we get:

Frequency = 340 m/s / 30m = 11.3 Hz

Therefore, the lowest two frequencies of the sound are 8.5 Hz and 11.3 Hz. This means that the sound waves emitted by the loudspeaker have two different frequencies, which are heard by the two listeners at different distances from the loudspeaker.

In summary, to find the lowest two frequencies of the sound, we need to use the formula frequency = speed of sound / wavelength and take into account the different distances between the loudspeaker and the listeners. I hope this helps in understanding how to approach this problem.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top