Calculate the frequency of the sound given the period:

\[ T = 2.3 \text{ msec}, \quad f = \]
\[ T = 0.005 \text{ sec}, \quad f = \]
\[ T = 25 \mu \text{sec}, \quad f = \]

Calculate the period given the frequency:

\[ f = 660 \text{ Hz}, \quad T = \]
\[ f = 100.5 \text{ MHz}, \quad T = \]
\[ f = 2.5 \text{ kHz}, \quad T = \]

3. A musician is playing a pan flute (blowing a pipe closed on one end). The length of the pipe is 15 cm (hint: the length \( L \) of the pipe is \( \frac{1}{4} \) of the wavelength).

\[ \text{What will be the frequency of the sound if the pipe is blown at 45°C?} \]
\[ \text{What will be the frequency of the sound if the pipe is blown at 5°C?} \]
\[ \text{To what conclusion do the results of questions a and b lead you?} \]

As the temperature drops, the frequency of the pan flute's sound also drops. This is due to the fact that when temperature drops, so does the sound speed in air.

1. Frequency calculation given the period

T = 2.3 msec, first

We are aware that frequency is the inverse of period. Therefore,

f=1/T, f=1/0.0023, and f=434.78 Hz

The sound has a 434.78 Hz frequency.

b. T = 0.005 sec

By applying the formula,

f=1/T, f=1/0.005, and f=200 Hz

The sound has a frequency of 200 Hz.

c. T = 25 µsec

It is stated in microseconds. Before computing the frequency, we must convert it to seconds.

T = 25 × 10⁻⁶ sec

By applying the formula,

f = 1/T

f = 1/(25 × 10⁻⁶)

f = 40,000 Hz

The sound has a frequency of 40,000 Hz.

2. Period calculation given frequency:

a. f = 660 Hz

By applying the formula,

T = 1/f

T = 1/660

T = 0.001515 sec

The The sound has a 0.001515 second period.

b. f = 100.5 MHz

The frequency is expressed in MHz. Prior to computing the period, we must convert it to Hz.

f = 100.5 × 10⁶ Hz

By applying the formula,

T = 1/f

T = 1/(100.5 × 10⁶)

T = 9.9256 × 10⁻⁹ sec

The sound has a duration of 9.9256 109 seconds.

c. f = 2.5 KHz

The frequency is expressed in KHz. Prior to computing the period, we must convert it to Hz.

f = 2.5 × 10³ Hz

By applying the formula,

T = 1/f

T = 1/(2.5 × 10³)

T = 0.0004 sec

The sound has a 0.0004 second period.

3. A pan flute's frequency may be calculated using the following information: the pipe is 15 cm long and closed on one end. The formula may be used to determine the sound's frequency.

f = (n × v)/(4L)

where L is the length of the pipe, v is the speed of sound in air, and n is the harmonic number (1, 2, 3,...).

When air is heated to 450 degrees Celsius, the speed of sound in the air changes. We may approximate using

v = 331 + 0.6T

where T is the Celsius temperature.

v = 331 + 0.6 × 45 v = 358.4 m/s

L = 15 cm = 0.15 m

When n = 1, the fundamental frequency,

f = (n × v)/(4L)

f = (1 × 358.4)/(4 × 0.15) f = 597.3 Hz

The pan flute produces sound at a frequency of 597.3 Hz.

v = 331 if the pipe is blasted at 50 degrees Celsius. + 0.6 × 5 v = 334 m/s

L = 15 cm = 0.15 m

When n = 1, the fundamental frequency,

f = (n × v)/(4L)

f = (1 × 334)/(4 × 0.15)

f = 556.7 Hz

Learn more about fundamental frequency from here;

https://brainly.com/question/31314205

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