Physics, 22.04.2020 02:09 mooreadrian412

Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds the other end loosely in his hand, so that the speed of waves on the clothesline is a relatively slow 0.740m/s . He finds several frequencies at which he can oscillate his end of the clothesline so that a light clothespin 43.0cm from the post doesn't move. What are these frequencies?

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Answer from: nicky123415

The obsrved frquencies are f= n (0.86) Hz n = 1,2,3,...,n.

Explanation:

From the question we are told that

The speed of the wave is v = 0.740 m/s

The distance from the post is $d = 43.0cm = \frac{43}{10} = 0.43m$

Generally frequency is mathematically represented as

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

so for the first frequency which the the fundamental frequency(first harmonic frequency) the clothesline(string) would form only one loop and hence the length between the vertical post and throcky hands is mathematically represented as

$L = \frac{\lambda}{2}$

=> $\lambda = 2L$

$f_f =f_1= \frac{v}{2L}$

Substituting values

$f_f = f_1 = \frac{0.740}{2* 0.43}$

= 0.860 Hz

For the second harmonic frequency i.e is when the clothesline(string) forms two loops the length is mathematically represented as

$L = \lambda$

So the second harmonic frequency is

$f_2 = \frac{v}{L}$

$f_2 = 2 * \frac{v}{2L}$

$f_2 =\frac{0.740}{0.430}$

= 1.72Hz

For the third harmonic frequency i.e when the clothesline(string) forms three loops the length is mathematically represented as

$L = 3 * \frac{\lambda }{2}$

So the wavelength is

$\lambda = \frac{2L}{3}$

And the third harmonic frequency is mathematically evaluated as

$f_3 = \frac{v}{\frac{2L}{3} }$

$= 3 * \frac{v}{2L}$

f_3=2.58Hz

Looking at the above calculation we can conclude that the harmonic frequency of a vibrating string(clothesline) can be mathematically represented as an integer multiple of the fundamental frequency

i.e

f = n f_f

$f = n \frac{v}{2L}$

f= n (0.86) Hz

Where n denotes integer values i.e n = 1,2,3,....,n.

Note : The length that exist between two nodes which are successive is equivalent to half of the wavelength so when one loop is form the number of nodes would be 2 and on anti-node

Answer from: Quest

answer: the correct answer id option(d).

explanation:

snell's law refraction:

$\frac{n_2}{n_1}=\frac{sin \theta_1}{sin\theta_2}$

the angle of refraction = $\theta_2=38 ^o$

the angle of incidence = $\theta_1=?$

refractive index of air = $n_1$ = 1

refractive index of glycerol $n_2$ = 1.47

$\frac{1.47}{1}=\frac{sin\theta_1}{sin 38^o}, (sin38^o=0.61)$

$sin\theta_2=0.8967$

$\theta_2=24.2^o, (sin^{-1}0.8967=63.72^o)$

from the given options, the most closest answer to 63.72° is 65° of option (d).

Answer from: Quest

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.

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