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3 years ago

If a violin string vibrates at 248 Hz as its fundamental frequency, what are the frequencies of the first four harmonics?

Physics

1 answer:

Andre45 [30]3 years ago

6 0

Answer:

The frequencies of the first four harmonics are 248 Hz, 496 Hz, 744 Hz, and 992 Hz

Explanation:

Given;

fundamental frequency, f₀ = 248 Hz

Frequency of the nth harmonic = n x fundamental frequency

First harmonic, f₁ = 1 x f₀ = 248 Hz

Second harmonic;

2f₀ = 2 x 248 Hz = 496 Hz

Third harmonic

3f₀ = 3 x 248 Hz = 744 Hz

Fourth harmonic

4f₀ = 4 x 248 Hz = 992 Hz

Therefore, the frequencies of the first four harmonics are 248 Hz, 496 Hz, 744 Hz, and 992 Hz

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Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

$\phi=\frac{4\pi}{5} *1.8$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

7 0

3 years ago

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

4 years ago

Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di

olga nikolaevna [1]

<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

☆...hope this helps...☆

_♡_mashi_♡_

6 0

2 years ago

The Earth's gravity keeps us from flying out to space. How can you explain that we have a stronger gravitational attraction to t

alexandr402 [8]

The sun is bigger, but has less mass than the earth

5 0

3 years ago

• ¿Qué rapidez tendrá un sonido de 150 Hz con una longitud de onda de 2 metros?

kobusy [5.1K]

Answer:

Speed = 300 m/s

Explanation:

Given the following data;

Frequency = 150 Hz

Wavelength = 2 meters

To find the speed of the wave;

Mathematically, the speed of a wave is given by the formula:

$Speed = wavelength * frequency$

Substituting into the formula, we have;

$Speed = 2 * 150$

Speed = 300 m/s

7 0

3 years ago