Ask question

Ask question

All categories

3 years ago

15

A fire engine is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the fire engin

e draws closer? The frequency increases, and the pitch decreases The frequency increases, and the pitch increases The frequency decreases, and the pitch decreases The frequency decreases, and the pitch increases

1 answer:

ValentinkaMS [17]3 years ago

7 0

Answer:

The frequency increases, and the pitch increases

Explanation:

Doppler's law of sound is applicable in such case when the observer or the sound source or both are moving relative to each other.

In such a case due to space-time constraint the waveform of the sound adjust themselves so as to obey the law of conservation of energy.

<u>The apparent frequency of the sound for the observer is given by:</u>

$f_o=(\frac{s+v_o}{s+v_s} )f$ ....................................(1)

where:

$f_o=$ observed frequency

$f=$ original source frequency

$s=$ speed of sound

$v_s=$ speed of source relative to the observer (taken negative when approaching towards the observer and vice-versa)

$v_o=$ speed of observer relative to the source (taken negative when moving away from the source and vice-versa)

<u>According to the given situation, eq. (1) becomes:</u>

$f_o=(\frac{s}{s-v_s} )f$

Since, $\frac{s}{s-v_s} >1$

Therefore

$f_o>f$

Pitch is very closely related to the frequency, it means that how fast is the amplitude of sound varying with time.

You might be interested in

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago

choose the correct answer 14: which of the following force follows the inverse square law of distance A: gravitaional force B: e

Delicious77 [7]

Answer:

C. both A and B

Explanation:

Sana nakatulong

8 0

3 years ago

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600

3241004551 [841]

Answer:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

$v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}$

Where:

$v_{o}$ - Initial speed of the rocket, in m/s.

$v_{ex}$ - Exhaust gas speed, in m/s.

$m_{o}$ - Initial total mass of the rocket, in kg.

$m$ - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

$m(t) = m_{o} + r\cdot t$

The initial total mass of the rocket is:

$m_{o} = 750\,kg$

The fuel consumption rate is:

$r = -\frac{600\,kg}{30\,s}$

$r = -20\,\frac{kg}{s}$

The function for the current total mass of the rocket is:

$m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t$

The speed function of the rocket is:

$v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}$

The speed of the rocket at given instants are:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

7 0

3 years ago

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

Licemer1 [7]

Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

Answer:

a

$\Delta f = 81.93 \ Hz$

b

$\lambda_1 = 0.867 \ m$

Explanation:

From the question we are told that

The speed of the train is $v_t = 39.6 m/s$

The frequency of the train horn is $f_t = 350 \ Hz$

Generally the speed of sound has a constant values of $v_s = 343 m/s$

Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_1 = f * \frac{v_s}{v_s - v_t}$

substituting values

$f_1 = 350 * \frac{343 }{343-39.6}$

$f_1 = 395.7 \ Hz$

Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_2 = f * \frac{v_s}{v_s +v_t}$

substituting values

$f_2 = 350 * \frac{343}{343 + 39.6}$

$f_2 = 313.77 \ Hz$

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

$\Delta f = f_1 - f_2$

$\Delta f = 395.7 - 313.77$

$\Delta f = 81.93 \ Hz$

Generally the wavelength detected by the person as the train approaches is mathematically represented as

$\lambda_1 = \frac{v}{f_1 }$

$\lambda_1 = \frac{343}{395.7 }$

$\lambda_1 = 0.867 \ m$

4 0

3 years ago

Students repeat the experiment but replace block X and block Y with block W and block Z , as shown in Figure 3. Block W and bloc

Lena [83]

The block Z would be seen in figure 10 when 4 strident turn around

8 0

3 years ago