Physics

Acoustics Issues


Sound and its spread

1. Sound is a mechanical wave that travels in the air at a variable speed, depending on the local temperature.

Assuming that in one place this speed is 340m / s. If a speaker vibrating its membrane at this location emits 1 250 pulses per second:

(a) determine the frequency of membrane vibration in Hertz;

This answer is found in the statement itself, since if the membrane emits 1,250 pulses per second, it repeats its movement 1,250 times per second, that is, its frequency.

b) Determine the period of vibration;

Knowing the frequency, we just need to remember that the period is equal to the inverse of the frequency, so:

Being the unit expressed in seconds that is the inverse unit the Hz.

(c) determine the wavelength of the sound wave in meters;

Using the equation:

We already know speed and frequency, so just isolate the wavelength:

d) Knowing that the velocity of sound in air varies with temperature according to the where θ is in degrees Celsius and the speed in meters per second. What is the local temperature?

Knowing that the speed of sound on site is 340m / s, we can use the equation and solve it:

2. Suppose in one place the speed of sound is 300m / s at a temperature of 0 ° C. In this same place temperatures during a certain time of the year can reach 40 ° C. At this temperature extreme what will be the speed of sound propagation?

Using the equation:

Where k is a constant of arbitrary value and T is the absolute ambient temperature. We can apply the values ​​to the equation in both situations:

and

Converting temperatures we have 273K and 313K respectively.

Dividing one equation by another:

Acoustic Interval

1. Two tuning forks are played at the same time. One has a frequency equal to 14kHz and the other 7kHz. What is the name of the acoustic interval between them?

Using the acoustic range equation we have:

Looking at a table, we find that the 2: 1 range is called octave.

2. A pair of sounds has a fifth acoustic range. Both sounds have the same propagation speed and the higher frequency sound has a wavelength of 1.3cm. What is the wavelength of the lowest frequency sound?

To solve this problem we must use the equation

Along with:

Which can be written as:

Joining the two equations:

Applying the known values, knowing that a fifth equals the 3: 2 quotient.