#2 in image 2 Trumpet PhysicsAs we learned in the video lesson, an open pipe of length L can create standingwaves at certain specific frequencies, given by:fn= n f12L= nfn = the frequency that creates a standing wave with n loops, in Hz= the "nth harmonic"fi = the frequency that creates a standing wave with 1 loop, in Hz= the "fundamental frequency"n = number of loops in the standing wavev = speed of the sound wave = 343 m/s (at 20° C)L= length of the pipe, in m%3DThe fundamental frequency, f1, is the lowest frequency that makes a standing wave onthe pipe. The higher frequencies or “harmonics" are multiples of the fundamental: f2is twice as much as f1; f3 is three times as big as f1, and so on.A trumpet is essentially a long, curved open pipe. (Actually, its geometry ismore complicated, but it is designed to produce the same set of frequencies that anopen pipe would. So close enough.) It has three valves, each of which is connected toan extra bit of tubing.When no valves are pushed, the sound wave passes straight through the horn,travelling a distance L. When a valve is pushed, it opens up the extra bit of tubing,which increases the length L of the horn slightly. According to the equation above,increasing the length L lowers the frequency of the standing wave, which produces alower tone or note. So, when a valve is pushed, it lowers the note the horn produces.Pushing a valve with a longer tube lowers the note even more. Opening more than onevalve makes the sound travel through more than one extra bit of tubing, lowering thenote even further. By picking the right combination of valves, the trumpet can makeany note on the musical scale.Directions: Here are some practice problems to make you familiar with theseconcepts. Please review the following. Assume the speed of sound is 343 m/s. 1. When no valves are pushed, the sixth harmonic of the horn (n = 6) produces afrequency of f6 = 698 Hz (an F). Use that information, and rearrange theequation above to calculate the length L of the trumpet's pipe when no valvesare used.2. When the first valve is pushed, the frequency of the sixth harmonic changes tofo = 622 Hz (an E-flat). Calculate the new length of the pipe L.3. The new length is slightly longer than the original length, because of the addedlength of the tubing attached to the first valve. Subtract the two lengths to findthe length of the tube attached to the first valve.4. When the second valve (only) is pushed, the frequency of the 6™ harmonic changesto fo = 659 Hz (an E). What is the new length of the pipe L?5. The difference in length between Step 4 and Step 1 is due to the length of the tubingattached to the second valve. Calculate the length of the tube attached to thesecond valve.6. When the third valve (only) is pushed, the frequency of the 6th harmonic changes tofo = 585 Hz (a D). What is the new length of the pipe L?7. The difference in length between Step 6 and Step 1 is due to the length of the tubingattached to the third valve. Calculate the length of the tube attached to the thirdvalve.8. The trumpet player now pushes both the 1" and 2nd valves, opening both bits ofextra tubing. What is the new length L of the horn?9. What is the frequency of the 6th harmonic produced in Problem 8?10. The trumpet player now pushes both the 1st and 3ª valves, opening both bits ofextra tubing. What is the new length L of the horn? What is the frequency ofthe 6th harmonic produced?11. The trumpet player now pushes both the 2nd and 3d valves, opening both bits ofextra tubing. What is the new length L of the horn? What is the frequency ofthe 6th harmonic produced?

Question

#2 in image 2 Trumpet PhysicsAs we learned in the video lesson, an open pipe of length L can create standingwaves at certain specific frequencies, given by:fn= n f12L= nfn = the frequency that creates a standing wave with n loops, in Hz= the &#34;nth harmonic&#34;fi = the frequency that creates a standing wave with 1 loop, in Hz= the &#34;fundamental frequency&#34;n = number of loops in the standing wavev = speed of the sound wave = 343 m/s (at 20° C)L= length of the pipe, in m%3DThe fundamental frequency, f1, is the lowest frequency that makes a standing wave onthe pipe. The higher frequencies or “harmonics&#34; are multiples of the fundamental: f2is twice as much as f1; f3 is three times as big as f1, and so on.A trumpet is essentially a long, curved open pipe. (Actually, its geometry ismore complicated, but it is designed to produce the same set of frequencies that anopen pipe would. So close enough.) It has three valves, each of which is connected toan extra bit of tubing.When no valves are pushed, the sound wave passes straight through the horn,travelling a distance L. When a valve is pushed, it opens up the extra bit of tubing,which increases the length L of the horn slightly. According to the equation above,increasing the length L lowers the frequency of the standing wave, which produces alower tone or note. So, when a valve is pushed, it lowers the note the horn produces.Pushing a valve with a longer tube lowers the note even more. Opening more than onevalve makes the sound travel through more than one extra bit of tubing, lowering thenote even further. By picking the right combination of valves, the trumpet can makeany note on the musical scale.Directions: Here are some practice problems to make you familiar with theseconcepts. Please review the following. Assume the speed of sound is 343 m/s. 1. When no valves are pushed, the sixth harmonic of the horn (n = 6) produces afrequency of f6 = 698 Hz (an F). Use that information, and rearrange theequation above to calculate the length L of the trumpet's pipe when no valvesare used.2. When the first valve is pushed, the frequency of the sixth harmonic changes tofo = 622 Hz (an E-flat). Calculate the new length of the pipe L.3. The new length is slightly longer than the original length, because of the addedlength of the tubing attached to the first valve. Subtract the two lengths to findthe length of the tube attached to the first valve.4. When the second valve (only) is pushed, the frequency of the 6™ harmonic changesto fo = 659 Hz (an E). What is the new length of the pipe L?5. The difference in length between Step 4 and Step 1 is due to the length of the tubingattached to the second valve. Calculate the length of the tube attached to thesecond valve.6. When the third valve (only) is pushed, the frequency of the 6th harmonic changes tofo = 585 Hz (a D). What is the new length of the pipe L?7. The difference in length between Step 6 and Step 1 is due to the length of the tubingattached to the third valve. Calculate the length of the tube attached to the thirdvalve.8. The trumpet player now pushes both the 1&#34; and 2nd valves, opening both bits ofextra tubing. What is the new length L of the horn?9. What is the frequency of the 6th harmonic produced in Problem 8?10. The trumpet player now pushes both the 1st and 3ª valves, opening both bits ofextra tubing. What is the new length L of the horn? What is the frequency ofthe 6th harmonic produced?11. The trumpet player now pushes both the 2nd and 3d valves, opening both bits ofextra tubing. What is the new length L of the horn? What is the frequency ofthe 6th harmonic produced?

Accepted Answer

Ques:

2. When the first valve is pushed, the frequency of the sixth harmonic changes to fo = 622 Hz (an E-flat). Calculate the new length of the pipe L.