In the previous section we saw the basic details about Longitudinal waves. In this section we will see some mathematical calculations involved with it.
Consider the compressions and rarefactions produced by a sound source shown in the fig.5.3(a) below:
It is clear that the molecules between A and M are under various levels of pressure. For example:
• Molecules that fall along the magenta coloured lines will be experiencing maximum compression. That is., maximum pressure
• Molecules that fall along the yellow coloured lines will be experiencing the least compression. That is., least pressure
• Other molecules which fall along neither of these lines will be experiencing intermediate pressures.
• Those which are closer to the magenta lines will be experiencing more pressure
• Those which are closer to the yellow lines will be experiencing less pressure
• So we find that, there is a variation of pressure along the path from the sound source to the ear
■ We can plot a graph with distance on the x-axis and pressure on the y-axis. Let us see how the plotting is done:
1. Consider the left most magenta line. Get the pressure experienced by the molecules along that line. Mark this pressure on the Y axis. This is point A.
2. Consider the left most yellow line. Get the pressure experienced by the molecules along that line.
♦ Mark this pressure on the Y axis.
♦ Draw a horizontal yellow dotted line through this mark
♦ Draw a vertical yellow dotted line from the left most yellow line
♦ The horizontal and vertical yellow dashed lines will intersect at a point C
♦ The x coordinate of C will give the distance from the sound source and the y coordinate will give the pressure.
3. So we successfully marked two points A and C
• y coordinate of A will give the pressure experienced by all molecules which fall along the left most magenta line
♦ x coordinate of A will be zero because, it's distance from the sound source is zero
• y coordinate of C will give the pressure experienced by all molecules which fall along the left most yellow line
♦ x coordinate of C will be the distance from the sound source
4. The intermediate points between A and C can be marked in the same way.
♦ y coordinate of those points will be the pressure,
♦ x coordinates will be the distance from the sound source.
5. Once we mark those points, we can join them. We will get a smooth curve between A and C
6. In this way, we can mark several points between C and M.
♦ A sample point at E is shown in the fig.5.3(a)
7. When we join all such points, we will get a smooth curve between A and M.
■ This curve is the graphical representation of the sound produced by the tuning fork.
8. We have to make a modification to this graph:
• Consider the pressure value at A. It is the maximum compression value
• Consider the pressure value at C. It is the least compression value
• Take the average of the two. From maths classes we know that, average of the two will be the exact midpoint between the following two points:
♦ The point A
♦ Projection of C on the y-axis. That is., the point of intersection of the horizontal yellow dotted line and the y axis
9. Mark this average point on the y axis.
10. Draw the x-axis through this average point. This is shown in fig.5.3(b)
♦ All points on the graph above the new x-axis represents a compression
♦ All points below the new x-axis represents a rarefaction.
■ This is the final graphical representation.
• Points B, D, F, H, J and L are the intersection points between the graph and the new x-axis.
• If we draw vertical lines through these points, the molecules which fall along those points will be experiencing the average pressure
2. Suppose that we are standing near the path of the wave, ready with a stop watch. Let the point where we are standing be 'P'
3. At the instant when a compression pass the point P, start the stop watch
4. Count the 'number of compressions' n, that pass the point P for any convenient time interval t, say 10 or 15 seconds.
5. Then the ratio n⁄t will give the 'number of compressions' that pass the point p in one second.
6. We know that, frequency is the 'number of occurrences of an event in unit time'
So n⁄t is the frequency of the wave. It is represented by ν (Greek letter nu). It's unit is hertz (symbol Hz).
• This name is given in honour of the German scientist Heinrich Rudolph Hertz, whose works laid the foundation of the future developments of radio, telegraph, telephone etc.,
2. At the instant when a compression pass the point P, start the stop watch
3. At the instant when the next compression pass the point P, stop the stop watch. Note down the time 'T'
4. This is the time taken between two consecutive compressions. It is called the 'time period' of the wave. It is represented by T. It's unit is seconds.
1. In the second experiment, we cannot take any convenient time
• We must stop the stop watch exactly at the instant at which the next compression pass the point P
2. So applying the result from the first experiment into the second, we have t = T
• Also, only one compression passes in a time of 'T'. So 'n' = 1
3. Thus we get: ν = n⁄t = 1⁄T
• We can write:
Eq.5.1: ν = 1⁄T
4. That means, frequency of a wave is the reciprocal of it's time period
5. 1⁄T has the unit s-1. So ν also has the unit s-1
6. That means, we can use either of the two units s-1 or Hz for the frequency ν
1. Mark two points at a convenient distance s apart. Say 150 or 250 m.
2. Note the time t1 when the car passes the first point
3. Note the time t2 when the car passes the second point
4. The difference (t2 – t1) will give the time t required by the car to travel s
5. So speed of the car = distance⁄time = s⁄t
■ The same method can be used to determine the speed of a train.
A train is a very long object. But we can take the times when the 'train's front end' passes the two points. The calculations are same as that of the car.
2. Two red arrows A and B are marked at the midpoint of two adjacent compartments. Let the distance between those two arrows be s. A green arrow C is marked on the platform.
3. Note the time t1 when the arrow B passes the arrow C. Note the time t2 when the arrow A passes the arrow C
4. The difference (t2 – t1) will give the time t required by the train to travel s
5. So speed of the train = distance⁄time = s⁄t
2. Then the distance s will correspond to λ
3. We know that, the time required for two consecutive crests to pass a fixed point such as the green arrow C is the time period T
4. So speed v of the wave can be obtained as: v = distance⁄time = λ⁄T
5. This can be written in another form also:
v = λ⁄T ⇒ v = λ × 1⁄T ⇒ v = λ × ν (∵ 1⁄T = ν)
We can write it in the form of an equation:
Eq.5.2:
v = λ × ν
That is., Speed of sound = wavelength × frequency
A sound wave has a frequency of 2 kHz and a wavelength of 35 cm. How long will it take to travel 1.4 km?
Solution:
■ We have, speed = wavelength × frequency
Given:
• Wave length, λ = 35 cm = 0.35 m
• Frequency, ν = 2 kHz = 2000 Hz = 2000 s-1
• So we get: v = 0.35 (m) × 2000 (s-1) = 700 m s-1
■ We have, speed, v = distance⁄time ⇒ time = distance⁄speed
Given:
• Distance = 1.4 km = 1400 m
• So time required to travel 1.4 km = 1400⁄700 = 2 s
Solved example 5.2
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m s-1 in a given medium
Solution:
■ We have, speed, v = λ × ν
So we can write: 440 = λ × 220 ⇒ λ = 440⁄220 = 2 m
Solved example 5.3
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Solution:
■ We have seen that the time interval between two consecutive compressions or rarefactions is the time period T of the wave. So in this problem we have to calculate T
■ We have also seen that frequency is the reciprocal of T. That is., ν = 1⁄T (Eq.5.1)
⇒ T = 1⁄ν
• So we get T = 1⁄500 = 0.002 seconds
In the next section, we will see some more properties of sound waves.
Consider the compressions and rarefactions produced by a sound source shown in the fig.5.3(a) below:
 |
| Fig.5.3 |
• Molecules that fall along the magenta coloured lines will be experiencing maximum compression. That is., maximum pressure
• Molecules that fall along the yellow coloured lines will be experiencing the least compression. That is., least pressure
• Other molecules which fall along neither of these lines will be experiencing intermediate pressures.
• Those which are closer to the magenta lines will be experiencing more pressure
• Those which are closer to the yellow lines will be experiencing less pressure
• So we find that, there is a variation of pressure along the path from the sound source to the ear
■ We can plot a graph with distance on the x-axis and pressure on the y-axis. Let us see how the plotting is done:
1. Consider the left most magenta line. Get the pressure experienced by the molecules along that line. Mark this pressure on the Y axis. This is point A.
2. Consider the left most yellow line. Get the pressure experienced by the molecules along that line.
♦ Mark this pressure on the Y axis.
♦ Draw a horizontal yellow dotted line through this mark
♦ Draw a vertical yellow dotted line from the left most yellow line
♦ The horizontal and vertical yellow dashed lines will intersect at a point C
♦ The x coordinate of C will give the distance from the sound source and the y coordinate will give the pressure.
3. So we successfully marked two points A and C
• y coordinate of A will give the pressure experienced by all molecules which fall along the left most magenta line
♦ x coordinate of A will be zero because, it's distance from the sound source is zero
• y coordinate of C will give the pressure experienced by all molecules which fall along the left most yellow line
♦ x coordinate of C will be the distance from the sound source
4. The intermediate points between A and C can be marked in the same way.
♦ y coordinate of those points will be the pressure,
♦ x coordinates will be the distance from the sound source.
5. Once we mark those points, we can join them. We will get a smooth curve between A and C
6. In this way, we can mark several points between C and M.
♦ A sample point at E is shown in the fig.5.3(a)
7. When we join all such points, we will get a smooth curve between A and M.
■ This curve is the graphical representation of the sound produced by the tuning fork.
8. We have to make a modification to this graph:
• Consider the pressure value at A. It is the maximum compression value
• Consider the pressure value at C. It is the least compression value
• Take the average of the two. From maths classes we know that, average of the two will be the exact midpoint between the following two points:
♦ The point A
♦ Projection of C on the y-axis. That is., the point of intersection of the horizontal yellow dotted line and the y axis
9. Mark this average point on the y axis.
10. Draw the x-axis through this average point. This is shown in fig.5.3(b)
♦ All points on the graph above the new x-axis represents a compression
♦ All points below the new x-axis represents a rarefaction.
■ This is the final graphical representation.
• Points B, D, F, H, J and L are the intersection points between the graph and the new x-axis.
• If we draw vertical lines through these points, the molecules which fall along those points will be experiencing the average pressure
Let us learn the salient features of this graph. The graph in fig.5.3(b) is shown again in fig.5.4 below:
• We can see that it contains some peak points and some valley points.
 |
| Fig.5.4 |
• The peak points are called crests of the wave. A, E, I and M are crests.
• Valley points are called troughs of the wave. C, G and K are troughs.
• The distance between two consecutive crests is called the wavelength. It is represented by λ (Greek letter lambda). The unit of wavelength is m.
Next we have to find the frequency of the wave. Let us see how it is calculated:
1. We have seen that the sound is propagated as compressions (represented by crests in the graph) and rarefactions (represented by troughs in the graph).2. Suppose that we are standing near the path of the wave, ready with a stop watch. Let the point where we are standing be 'P'
3. At the instant when a compression pass the point P, start the stop watch
4. Count the 'number of compressions' n, that pass the point P for any convenient time interval t, say 10 or 15 seconds.
5. Then the ratio n⁄t will give the 'number of compressions' that pass the point p in one second.
6. We know that, frequency is the 'number of occurrences of an event in unit time'
So n⁄t is the frequency of the wave. It is represented by ν (Greek letter nu). It's unit is hertz (symbol Hz).
• This name is given in honour of the German scientist Heinrich Rudolph Hertz, whose works laid the foundation of the future developments of radio, telegraph, telephone etc.,
Let us repeat the experiment. This time, to determine another quantity.
1. Suppose that we are standing near the path of the wave, ready with a stop watch. Let the point where we are standing be 'P'2. At the instant when a compression pass the point P, start the stop watch
3. At the instant when the next compression pass the point P, stop the stop watch. Note down the time 'T'
4. This is the time taken between two consecutive compressions. It is called the 'time period' of the wave. It is represented by T. It's unit is seconds.
In the first experiment also, we did a similar procedure. Is there any inter connection?
Let us see:1. In the second experiment, we cannot take any convenient time
• We must stop the stop watch exactly at the instant at which the next compression pass the point P
2. So applying the result from the first experiment into the second, we have t = T
• Also, only one compression passes in a time of 'T'. So 'n' = 1
3. Thus we get: ν = n⁄t = 1⁄T
• We can write:
Eq.5.1: ν = 1⁄T
4. That means, frequency of a wave is the reciprocal of it's time period
5. 1⁄T has the unit s-1. So ν also has the unit s-1
6. That means, we can use either of the two units s-1 or Hz for the frequency ν
Speed of a wave
We know how to find the speed of a car. Let us recall:1. Mark two points at a convenient distance s apart. Say 150 or 250 m.
2. Note the time t1 when the car passes the first point
3. Note the time t2 when the car passes the second point
4. The difference (t2 – t1) will give the time t required by the car to travel s
5. So speed of the car = distance⁄time = s⁄t
■ The same method can be used to determine the speed of a train.
A train is a very long object. But we can take the times when the 'train's front end' passes the two points. The calculations are same as that of the car.
■ For the train, another method can also be used:
1. Consider fig.5.5 below. It shows a train moving from left to right.  |
| Fig.5.5 |
3. Note the time t1 when the arrow B passes the arrow C. Note the time t2 when the arrow A passes the arrow C
4. The difference (t2 – t1) will give the time t required by the train to travel s
5. So speed of the train = distance⁄time = s⁄t
The same method can be used to find the speed of a wave.
1. The two arrows A and B correspond to two consecutive crests of a wave2. Then the distance s will correspond to λ
3. We know that, the time required for two consecutive crests to pass a fixed point such as the green arrow C is the time period T
4. So speed v of the wave can be obtained as: v = distance⁄time = λ⁄T
5. This can be written in another form also:
v = λ⁄T ⇒ v = λ × 1⁄T ⇒ v = λ × ν (∵ 1⁄T = ν)
We can write it in the form of an equation:
Eq.5.2:
v = λ × ν
That is., Speed of sound = wavelength × frequency
Now we will see a solved example
Solved example 5.1A sound wave has a frequency of 2 kHz and a wavelength of 35 cm. How long will it take to travel 1.4 km?
Solution:
■ We have, speed = wavelength × frequency
Given:
• Wave length, λ = 35 cm = 0.35 m
• Frequency, ν = 2 kHz = 2000 Hz = 2000 s-1
• So we get: v = 0.35 (m) × 2000 (s-1) = 700 m s-1
■ We have, speed, v = distance⁄time ⇒ time = distance⁄speed
Given:
• Distance = 1.4 km = 1400 m
• So time required to travel 1.4 km = 1400⁄700 = 2 s
Solved example 5.2
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m s-1 in a given medium
Solution:
■ We have, speed, v = λ × ν
So we can write: 440 = λ × 220 ⇒ λ = 440⁄220 = 2 m
Solved example 5.3
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Solution:
■ We have seen that the time interval between two consecutive compressions or rarefactions is the time period T of the wave. So in this problem we have to calculate T
■ We have also seen that frequency is the reciprocal of T. That is., ν = 1⁄T (Eq.5.1)
⇒ T = 1⁄ν
• So we get T = 1⁄500 = 0.002 seconds
In the next section, we will see some more properties of sound waves.
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