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

Answer from: 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.

Answer from: Quest

answer: he got 23

explanation:

Answer from: Quest

0.19 m/s and 0.45 m/s

Another question on Physics

Physics, 21.06.2019 17:10

Which statement about covalent bonds is true

Answers: 1

Answer

Physics, 21.06.2019 23:00

1point) a skydiver weighing 130 lb (including equipment) falls vertically downward from an altitude of 14,000 ft and opens the parachute after 20 s of free fall. assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude 0.85|v|0.85|v| when the parachute is closed and is of magnitude 17|v|17|v| when the parachute is open, where the velocity vv is measured in ft/s. assume that acceleration due to gravity has magnitude 32 ft/s/s; remember that weight is the product of mass and gravitational acceleration. find the speed of the skydiver when the parachute opens. speed is 150.6121788 ft/s find the skydiver's height when the parachute opens. height is 2338.98591 ft find the limiting velocity vlvl after the parachute opens. (think about whether this should be positive or negative.) vl=vl= 7.647 ft/s

Answers: 2

Answer

Physics, 22.06.2019 09:50

A250-n box is at rest on a frictionless horizotal floor. then a person shoves the box with a non-zero force of magnitude p. which statement best describes the relationship betweeen p and the motion of the box after it was shoved? -the box will not move when p< 250n but otherwise will accelerate at a constant rate that is directly proportional to p - the box will begin moving at a speed that is directly proportional to p, then will gradually slow to a halt -the box will move with constant velocity at a speed that is directly proportional to p -box will accelerate across the floor at a constant rate that is directly proportional to p

Answers: 1

Answer

Physics, 22.06.2019 22:20

Which symbol in the first law of thermodynamics represents the sum of the chemical and thermal energy stored in atoms and molecules? a. q b.w c.v d.u

Answers: 3

Answer

You know the right answer?

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

Questions