Physics, 04.01.2021 16:20 talia43

Standing waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under the same tension The lengths of the strings are given. Rank the frequencies of the oscillations, from largest to smallest

Standing waves are created in the four strings shown in Figure 25. All strings have the same mass p

Answers: 1

Show answers

Answers

Answer from: kaylaamberd

The order of the frequencies is f_Df_Bf_Af_C

Standing waves:

The frequency of the standing wave in a string tied at both ends is given by:

$f=\frac{nv}{2L}$

where n is the mode of frequency

v is the velocity of the wave

and L is the length of the string.

Now the velocity of a wave in a string tied at both ends is given by

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

where T is the tension and μ is the mass per unit length.

Since T and μ are the same for all the strings, velocity will be the same for all.

Now to find the mode of frequency we can calculate the number of nodes (including the nodes at the ends) in the given figure and subtract by 1. Nodes are the point where the amplitude of the wave is zero.

$f_A=\frac{3v}{2\times27}=\frac{v}{18}\;s^{-1}\\\\f_B=\frac{4v}{2\times30}=\frac{v}{15}\;s^{-1}\\\\f_C=\frac{3v}{2\times30}=\frac{v}{20}\;s^{-1}\\\\f_D=\frac{4v}{2\times24}= \frac{v}{12}\;s^{-1}$

Hence, f_Df_Bf_Af_C

Learn more about standing waves:

link

Answer from: ghostboy2001

The rank of the frequencies from largest to smallest is

The largest frequency of oscillation is given by the string in option D

The second largest frequency of oscillation is given by the string in option B

The third largest frequency of oscillation is given by the string in option A

The smallest frequency of oscillation is given by the string in option C

Explanation:

The given parameters are;

The mass per unit length of all string, m/L = Constant

The tension of all the string, T = Constant

The frequency of oscillation, f, of a string is given as follows;

$f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}$

Where;

T = The tension in the string

m = The mass of the string

L = The length of the string

n = The number of overtones

$Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K$

For the string in option A, the length, L = 27 cm, n = 3 we have;

$f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K$

For the string in option B, the length, L = 30 cm, n = 4 we have;

$f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K$

For the string in option C, the length, L = 30 cm, n = 3 we have;

$f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K$

For the string in option D, the length, L = 24 cm, n = 4 we have;

$f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K$

Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;

1 ) f_D 2) f_B 3) f_A 4) f_C

Answer from: Quest

electrical power

Answer from: Quest

No bc a lot of things r still theories n unproven

Other questions on the subject: Physics

Physics, 21.06.2019 22:00, moneybaggzay123

1. consider the case in which air fills air shocks on a truck trailer. the pressure in the shocks is 2 mpa. the temperature is 300 k. the diameter of the shock piston is 10 cm and the initial length of the cylindrical cavity containing the compressed air is 40 cm. a. the truck is gradually loaded over a period of a day in a static setting. the temperature is held constant for the atmosphere and thus for the gas shock. calculate the compressibility of the air in the shock for this condition when the truck is initially being loaded. b. if the shocks were loaded in a dynamic setting by driving over bumps, what would be the compressibility? state your assumption. c. what is the initial load on the shock if the shock is in an atmospheric 100 kpa? d. if the shock is compressed using the process described in part a, and the air shock compressed air cavity length decreases to 20 cm, what is the additional load applied to the shock?

Answers: 2

continue

Physics, 22.06.2019 00:40, naomicervero

The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if each bar has a cross-sectional area of 450 mm2, determine the maximum value of p if the allowable stress is (σallow)st = 200 mpa for the steel and ( σallow)al = 150 mpa for the aluminum. est = 200 gpa and eal = 70 gpa.

Answers: 1

continue

Physics, 22.06.2019 09:00, montgomeryaevans

This is really important 1.which of the following prefixes represents the largest value? (2 points)gigahectorkilomilli2.which of the following types of graphs is best for plotting the mean, median, and mode of data? (2 points)bar graphbox-and-whiskercircle graphstem-and-leaf

Answers: 1

continue

Physics, 23.06.2019 02:30, pruittjr5161

Based on its type of chemical bond, which of the following has the highest boiling point? potassium chloride ethyl alcohol water ammonia

Answers: 1

continue

You know the right answer?

Standing waves are created in the four strings shown in Figure 25. All strings have the same mass pe...