The fundamental frequency of a closed tube is the same frequency as the 2nd harm

Question

The fundamental frequency of a closed tube is the same frequency as the 2nd harmonic of a stretched string. The mass per unit length string is 6.10 x 10^-4 kg/m. The tension in the string is 48.0 N.

A) What is the ration of the length of the string to the length of the closed tube Ls/Lt?

(Write your answer as a number. for example if Ls=1 and Lt=2 write 0.5 since 1/2=0.5)

The length of the closed tube is 19.6 cm. The closed tube is played on one side of a room. The string is plucked on the other side of a room. A person jogs at a speed of 2m/s away from the closed tube and towards the string.

B) What beat frequency will the person hear due to interference between the fundamental frequancy of the closed tube and the second harmonic of the string?

Thank you so much!!!!

Explanation / Answer

The mass per unit length string is 6.10 x 10^-4 kg/m. The tension in the string is 48.0 N.

Speed of the wave V = (T ) = sqrt((48.0/6.10×10^-4) = 280.51 m/s

Frequency of string: f = nv/2Ls and Frequency of closed tube: f = nv/4Lt (suffix s for string length and t for tube length and n stand for harmonics, n=1 fundamental harmonics) since their frequnecy is same and n is diferent as per question

2*v/2Ls= 1*v/4Lt where n=2 for string and 1 for tube as per question)

=>2*280.51/Ls=1*343/2Lt

=>Ls/Lt=(2*280.51*2)/(1*343)=3.2713

Second part

The length of the closed tube is 19.6 cm. The closed tube is played on one side of a room. The string is plucked on the other side of a room. A person jogs at a speed of 2m/s away from the closed tube and towards the string.

B) What beat frequency will the person hear due to interference between the fundamental frequancy of the closed tube and the second harmonic of the string?

Frequency of closed tube=nv/4L=1*343/4*0.196= 437.5 Hz

Length of string =3.2713*0.196= 0.6412m =64.1 cm

frequency in string=nv/2L=2*280.51/2*0.641= 437.61 Hz

Beats occur whenever two sound sources emit sounds of slightly different frequencies. The beat frequency is just the difference in frequency of the two sources

Thus 437.61 -437.5 = 0.11 Hz actual beat freq

However, person relative velocity is 2 m/s

When the listener is moving away from A then the apparent frequency of sound is less than actual frequency of sound

n'=n*(343-u)/343=437.5*(343-2)/343= 434.9490 hz ( here 343 m/s velcoity of sound in air)

similarly, when the listener is approaching B then the apparent frequency of sound  

n'=437.61*(343+u)/343= 440.1617 hz

apparent beat he will hear= 440.1617 - 434.9490=5.2127 hz i.e appx 5 beat per second person will hear

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.