## 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.*