This is the reduction in the density of a medium

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

so, the difference between any two harmonics tube is equal to:

$f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 4699 Hz\\f_{n+1}=4953 Hz$

So, according to (1), the fundamental frequency is equal to half of this difference:

$f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz$

Now we know that one of the harmonics is $f_n=4191 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz$

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

Since we know $f_n = 4445 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17$

7 0

3 years ago

A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potentia

den301095 [7]

Answer:

V = 575.6 Volts

Explanation:

As we know that surface area of the equi-potential surface is given as

$A = 1.18 m^2$

so we will say

$A = 4\pi r^2$

$1.18 = 4\pi r^2$

$r = 0.31 m$

Now the potential due to a point charge is given as

$V = \frac{kQ}{r}$

$V = \frac{(9\times 10^9)(1.96 \times 10^{-8})}{0.31}$

$V = 575.6 Volts$

8 0

3 years ago

A piece of cork (density 250 kg/m3 ) of mass 0.01 kg is held in place under water (density 1000 kg/m3 ) by a string. What is the

vekshin1

Answer:

0.3 N

Explanation:

mass of cork = 0.01 kg, density of cork = 250 kg/m^3

density of water = 1000 kg/m^3, g = 10 m/s^2

Tension in the rope = Buoyant force acting on the cork - Weight of the cork

Buoyant force = volume of cork x density of water x g

= mass x density of water x g / density of cork

= 0.01 x 1000 x 10 / 250 = 0.4 N

Weight of cork = mass of cork x g = 0.01 x 10 = 0.1 N

Thus, the tension in the rope = 0.4 - 0.1 = 0.3 N

3 0

3 years ago

Read 2 more answers

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 460 nm occurs at an angl

Svet_ta [14]

Answer:

a)λ = 6.133 10⁻⁷ m = 613.3 nm, b) d = 1.60 10⁻⁶ m , c) Δd = 0.24 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon for the double slit is

d sin θ = m λ m = ±1, ±2,…

a) Let's use the data to find the separation of the slits (d)

d = m λ / sin θ

d = 4 460 10⁻⁹ / sin 90

d = 1.84 10⁻⁶ m

Now we can calculate the minimum wavelength for the third orcen (m = 3), the maximum angle must be 90 °

λ = d sin θ / m

λ = 1.84 10⁻⁶ sin 90 / 3

λ = 6.133 10⁻⁷ m

This is the last visible wavelength in this interference order, that is, we see the light with a shorter length than the calculated

b) To eliminate all the light of the fourth order (m = 4) let's use the minimum wavelength of visible range ( λ = 400 nm) from a 90 ° angle

d = m λ / sin θ

d = 4 400 10⁻⁹ / sin 90

d = 1.60 10⁻⁶ m

If we reduce the slit to this value, the spectrum for the fourth order can be used.

c) It is change of separation is

Δd = d₂ - d₁

Δd = (1.84 - 1.60) 10⁻⁶ m

Δd = 0.24 10⁻⁶ m

6 0

3 years ago

The string is fixed at two ends with distance 1.5 m. Its mass is 5 g and the tension in the string is 50N and it vibrates on its

liraira [26]

Answer:

a) $\lambda=1\ m$

b) $f=122.47\ Hz$

c) $\lambda_s=2.8\ m$

Explanation:

Given:

distance between the fixed end of strings, $l=1.5\ m$

mass of string, $m=5\ g=0.005\ kg$

tension in the string, $F_T=50\ N$

a)

Since the wave vibrating in the string is in third harmonic:

Therefore wavelength λ of the string:

$l=1.5\lambda$

$\lambda=\frac{1.5}{1.5}$

$\lambda=1\ m$

b)

We know that the velocity of the wave in this case is given by:

$v=\sqrt{\frac{F_T}{\mu} }$

where:

$\mu=$ linear mass density

$v=\sqrt{\frac{50}{(\frac{m}{l}) } }$

$v=\sqrt{\frac{50}{(\frac{0.005}{1.5}) } }$

$v=122.47\ m.s^{-1}$

Now, frequency:

$f=\frac{v}{\lambda}$

$f=\frac{122.47}{1}$

$f=122.47\ Hz$

c)

When the vibrations produce the sound of the same frequency:

$f_s=122.47\ Hz$

Velocity of sound in air:

$v_s=343\ m.s^{-1}$

Wavelength of the sound waves in air:

$\lambda_s=\frac{v_s}{f_s}$

$\lambda_s=2.8\ m$

5 0

3 years ago