Physics, 13.01.2021 15:20, sheilashair

A whistle emitting a note of 4 kHz is whirled in a horizontal circle of radius 1 m at a constant speed. If the highest note heard
by a person a large distance away is 4200 Hz, find:
(i) the speed of the whistle
(ii) the lowest note heard by the person,
(iii) the time taken for the whistle to make on complete revolution,
(iv) the time interval between the person hearing the highest and lowest note.

Answers: 3

Show answers

Answers

answered: Alexandragurule18

(i) The speed of the whistle is 16. $\overline 3$ m/s

(ii) The lowest note heard by the person is 3,818. $\overline{18}$ Hz

(iii) The time taken for the whistle to make one revolution is approximately 0.38468 seconds

(iv) The time interval between the person hearing the highest and lowest note is approximately 0.19234 seconds

Explanation:

The question relates to the Doppler effect for which the change in frequency is given as follows;

$f_L = \left ( \dfrac{v \pm v_L }{v \pm v_S} \right ) \cdot f_S$

Where;

f_L = The frequency heard by the listener

v = The speed of sound in air = 343 m/s

v_L = The listener's speed

v_S = The source's (whistle) speed

f_S = The frequency of the source

The given parameters are;

The frequency of the whistle (source), f_S = 4 kHz = 4,000 Hz

The radius of the horizontal circle around which the whistle is whirled = 1 m

The highest note heard by a person a large distance away = 4200 Hz

The distant listener is taken as stationary, v_L = 0 m/s

We therefore have;

$f_L = \left ( \dfrac{v \pm 0 }{v \pm v_S} \right ) \cdot f_S = \left ( \dfrac{v }{v \pm v_S} \right ) \cdot f_S$

(i) When the whistle's motion while it is being whirled towards the listener, v_S is negative

When the whistle's motion while it is being whirled is away from listener, v_S is positive

The highest note, $f_{LH}$ , is obtained with the lowest denominator, that is when the source moves towards the listener and v_S is negative, we get;

$f_{LH} = 4,200 \ Hz = \left ( \dfrac{343 \ m/s }{343 \ m/s - v_S} \right ) \times 4,000 \ Hz$

$\dfrac{4,200 \ Hz}{4,000 \ Hz} = \dfrac{21}{20} = \dfrac{343 \ m/s}{343 \ m/s - v_S}$

21·(343 m/s - v_S ) = 20×343 m/s

21 × v_S = 21 × 343 m/s - 20 × 343 m/s = 343 m/s

v_S = (343 m/s)/21 = 49/3 m/s = 16. $\overline 3$ m/s

The speed of the source = The speed of the whistle

∴ The speed of the source = v_S = 16. $\overline 3$ m/s = The speed of the whistle

(ii) The lowest note heard by the person occurs when the denominator of the formula for the change in frequency is smallest for which we have the value of v_S as positive and the source moves away from the observer which is presented as follows;

$f_{LH} = \left ( \dfrac{343 \ m/s }{343 \ m/s +16.\overline 3 \ m/s } \right ) \times 4,000 \ Hz = 3,818.\overline {18} \ Hz$

The lowest note heard by the person = 3,818. $\overline{18}$ Hz

(iii) The angular velocity of the whistle, ω is given as follows;

$\omega = \dfrac{\theta}{t} = \dfrac{v_S}{r} = \dfrac{16. \overline 3 \ m/s}{1 \ m} = 16. \overline 3 \ radians/second$

The angle rotated in one revolution, is 2·π radians, therefore, when θ = 2·π radians, the time, t_c , to complete one revolution is given as follows;

$t_c = \dfrac{\theta}{\omega} = \dfrac{2 \times \pi \ radians}{16. \overline 3 \ radians/second} \approx 0.385 \ seconds$

Therefore, the time to complete one revolution ≈ 0.38468 seconds

(iv) The time interval between the person hearing the highest and lowest note is the time to complete half a cycle ≈ 0.38468/2 seconds ≈ 0.19234 seconds.

answered: Guest

i need the answer to this too but idk it ; -;

Predict how the weather in chicago, il will change in the next 12 hours. a. it will experience clea

answered: Guest

the simple machine has three main features. they are:

loadfulcrumeffort

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

Other questions on the subject: Physics

Physics, 22.06.2019 00:00, hlhunter11p8f0o0

An object moves from the position +34m to the position -15m in 15 seconds. what is it's total displacement? what is it's average velocity?

Answers: 1

continue

Physics, 22.06.2019 12:20, ginachuquiano450

Aball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. if the ball hits the floor with a speed of 2.40 m/s and rebounds with a speed of 1.70 m/s, determine the following. (a) magnitude of the change in the ball's momentum (let up be in the positive direction.)

Answers: 1

continue

Physics, 22.06.2019 21:10, tednequamoore4307

Aspecified volume of space contains an electric field for which the magnitude is given by e=e0cos(ωt). suppose that e0 = 20 v/m and ω = 1.0 × 107 s−1. part a what is the maximum displacement current through a 0.80 m2 cross-sectional area of this volume? express your answer with the appropriate units. idisp, max i d i s p , m a x = nothing nothing request answer part b how would this area have to be oriented relative to the electric field to get this maximum displacement current?

Answers: 2

continue

Physics, 23.06.2019 01:00, christopherwilde

Is a volume of space that is empty of matter.

Answers: 3

continue

Do you know the correct answer?

A whistle emitting a note of 4 kHz is whirled in a horizontal circle of radius 1 m at a constant sp...