When air is blown into the open pipe,
L =
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ= =3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f=
⇒f=
≈115 Hz
Answer:
It is producing either a 435-Hz sound or a 441-Hz sound.
Explanation:
When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:
(1)
In this problem, we know that:
- The frequency of the first trombone is
- 6 beats are heard every 2 seconds, so the beat frequency is
If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:
Answer:
Explanation:
Given
Initial velocity of first billiard ball
Initial velocity of second billiard ball
After collision first ball comes to rest
suppose m is the mass of both the balls
Conserving momentum to get the speed of second ball after collision
Initial momentum
Final momentum
where and
are the speed of first and second ball respectively
thus speed of second ball after collision is equal to speed of first ball before collision