1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
jeka57 [31]
3 years ago
14

An organ pipe is 248 cm long.(a) Determine the fundamental overtone if the pipe is closed at one end.(b) Determine the first aud

ible overtone if the pipe is closed at one end.(c) Determine the second audible overtone if the pipe is closed at one end.(d) Determine the third audible overtone if the pipe is closed at one end.(e) Determine the fundamental overtone if the pipe is open at both ends.(f) Determine the first audible overtone if the pipe is open at both ends.(g) Determine the second audible overtone if the pipe is open at both ends.(h) Determine the third audible overtone if the pipe is open at both end.
Physics
1 answer:
Phantasy [73]3 years ago
7 0

(a) 34.6 Hz

The fundamental frequency of a pipe closed at one end is given by

f_1 = \frac{v}{4 L}

where

v = 343 m/s is the speed of the sound in air

L is the length of the pipe

In this problem,

L = 248 cm = 2.48 m

So, the fundamental frequency is

f_1 = \frac{343 m/s}{4 (2.48 m)}=34.6 Hz

(b) 103.8 Hz

In a open-closed pipe, only odd harmonics are produced; therefore, the frequency of the first overtone (second harmonic) is given by:

f_2 = 3 f_1

where

f_1 = 34.6 Hz is the fundamental frequency.

Substituting into the equation,

f_2 = 3 (34.6 Hz)=103.8 Hz

(c) 173 Hz

The frequency of the second overtone (third harmonic) is given by:

f_3 = 5 f_1

where

f_1 = 34.6 Hz is the fundamental frequency.

Substituting into the equation,

f_3 = 5 (34.6 Hz)=173 Hz

(d) 242.2 Hz

The frequency of the third overtone (fourth harmonic) is given by:

f_4 = 7 f_1

where

f_1 = 34.6 Hz is the fundamental frequency.

Substituting into the equation,

f_4 = 7 (34.6 Hz)=242.2 Hz

(e) 69.2 Hz

The fundamental frequency of a pipe open at both ends is given by

f_1 = \frac{v}{2 L}

where

v = 343 m/s is the speed of the sound in air

L is the length of the pipe

In this problem,

L = 248 cm = 2.48 m

So, the fundamental frequency is

f_1 = \frac{343 m/s}{2 (2.48 m)}=69.2 Hz

(f) 138.4 Hz

In a open-open pipe, both odd and even harmonics are produced; therefore, the frequency of the first overtone (second harmonic) is given by:

f_2 = 2 f_1

where

f_1 = 69.2 Hz is the fundamental frequency.

Substituting into the equation,

f_2 = 2 (69.2 Hz)=138.4 Hz

(g) 207.6 Hz

The frequency of the second overtone (third harmonic) in an open-open pipe is given by:

f_3 = 3 f_1

where

f_1 = 69.2 Hz is the fundamental frequency.

Substituting into the equation,

f_3 = 3 (69.2 Hz)=207.6 Hz

(h) 276.8 Hz

The frequency of the third overtone (fourth harmonic) in an open-open pipe is given by:

f_4 = 4 f_1

where

f_1 = 69.2 Hz is the fundamental frequency.

Substituting into the equation,

f_4 = 4 (69.2 Hz)=276.8 Hz

You might be interested in
4. A hockey puck with a momentum of -17 kg x m/s when it
Nonamiya [84]

Answer:

-2 kg x m/s

Explanation:

-17 + 54 = 35 + x

x = -2

7 0
3 years ago
A car was moving a 14 m/s. After 30 seconds, it’s speed increased to 20 m/s. What was its acceleration during this time?
Anuta_ua [19.1K]

Answer:

acceleration = 0.2 m/s/s

Explanation:

initial velocity u = 14

final velocity v = 20

time = 30

acceleration = ?

v = u + at

20 = 14 + 30a

30a = 6

a = 0.2 m/s/s

5 0
2 years ago
Read 2 more answers
what is the amplitude and wavelength of the wave shown below? A. Amplitude: 15 cm; wavelength: 150 cm B. Amplitude: 30 cm; wavel
ehidna [41]
I would say C...............
5 0
3 years ago
Read 2 more answers
We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength
san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

\theta=1.22\frac{\lambda}{b}

θ = 50.0*10^-7 rad

λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm

hence, the smallest diameter objective lens is 13.4cm

8 0
3 years ago
Read 2 more answers
The pressure of air pushing down on a house roof is extremely large. Why is the roof not crushed?
Degger [83]
There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.
7 0
3 years ago
Other questions:
  • Which one is it ASAP
    12·1 answer
  • Which of the following is NOT NECESSARILY a property of an air mass?
    12·1 answer
  • When the Voyager I and Voyager II spacecraft were exploring the outer planets, NASA flight controllers had to plan the crafts' m
    13·1 answer
  • A diver shines a flashlight upward from beneath the water (n=1.33) at a 36.2° angle to the vertical. At what angle does the ligh
    5·1 answer
  • A stagehand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N. W
    10·1 answer
  • 6. In an integrated circuit, each wafer is cut into sections, which
    8·1 answer
  • A hole of radius r has been drilled in a circular, flat plate of radius R. The center of the hole is a distance d from the cente
    8·1 answer
  • I’m in the middle of a test right now. If anybody knows this can they help
    5·1 answer
  • 11. A roller coaster car is moving slowly up a long ramp to the very top of the roller coaster. It goes over the top and starts
    6·1 answer
  • A pulley system is used in an auto repair shop to lift a 200 kg engine block 1.5 m out of a car. The lower pulley is attached di
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!