The Student Room Group

Physics waves questions

Hi,soo sorry I have two more questions :frown:. For 2b, I would have thought it's the third harmonic so there'll be a node 1/3 of the distance, but the answer is 1/4 distance and I dotn really understand why.

Also, for question 3b, i dont get the path difference ting. why do you have to divide it by two, also we were never taught about path difference?

Thank you sooo much xxxx
hey so

a) Musical concert pitch has a frequency of 440 Hz.

A correctly tuned A-string on a guitar has a first harmonic (fundamental frequency) two
octaves below concert pitch.
Determine the first harmonic of the correctly tuned A-string.
frequency 110 HzHz

(b) Describe how a note of frequency 440 Hz can be produced using the correctly tuned Astring of a guitar.

(Use finger on the fret so that) a ¼ length of the string is used to sound the note or hold string down on 24th fret

(c) Describe the effect heard when notes of frequency 440 Hz and 430 Hz of similar
amplitude are sounded together.

Mention or description of beats or description of rising and falling amplitude / louder

and quieter
Regular rising and falling of loudness owtte

Beat frequency 10(.0Hz) Allow beat frequency = 430 - 420
Figure 1 and Figure 2 show a version of Quincke’s tube, which is used to demonstrate interference of sound waves. A loudspeaker at X produces sound waves of one frequency. The sound waves enter the tube
and the sound energy is divided equally before travelling along the fixed and movable tubes. The
two waves superpose and are detected by a microphone at Y.
(a) The movable tube is adjusted so that d1 = d2 and the waves travel the same distance from
X to Y, as shown in Figure 1. As the movable tube is slowly pulled out as shown in Figure
2, the sound detected at Y gets quieter and then louder.


Explain the variation in the loudness of the sound at Y as the movable tube is slowly pulled out.

Initially the path difference is zero/the two waves are in phase when they meet/the (resultant) displacement is a maximum.
As the movable tube is pulled out, the path difference increases and the two waves are no longer in phase, so the displacement and loudness decrease.When the path difference is one half wavelength, the two are in antiphase and sound is at its quietest.As the path difference continues to increase, the two waves become more in phase and the sound gets louder again.
The tube starts in the position shown in Figure 1.
Calculate the minimum distance moved by the movable tube for the sound detected at Y to
be at its quietest.
frequency of sound from loud speaker = 800 Hz
speed of sound in air = 340 m s–1


Use of wavelength = speed / frequency
To give: 340 / 800 = 0.425 m
Path difference = one half wavelength = 0.21 m
Path difference = 2 (d2– d1) = 2 (distance moved by movable tube)
Distance moved by movable tube = 0.10 m.

Quincke’s tube can be used to determine the speed of sound.
State and explain the measurements you would make to obtain a value for the speed of
sound using Quincke’s tube and a sound source of known frequency,


Start with d1 = d2
Measure distance moved by movable tube for each successive minima and maxima
Start with d1 = d2
Measure distance moved by movable tube for first minimum.
Each change in distance is equal to one quarter wavelength.
Distance is equal to one quarter wavelength
Continue until tube is at greatest distance or repeat readings for decreasing distance
back to starting point.
Repeat for different measured frequencies.
Use speed = frequency x wavelength
all done hope it helps!!
how do you get 0.1
how do you get 0.1m

Quick Reply

Latest