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kobusy [5.1K]
3 years ago
13

A 6.0 m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is 0.40

m, and the amplitude is 4.0 mm, is propagating on the wire. a) How long will it take for a crest of this wave to travel the length of the wire? b) What is the tension in the wire?
Physics
1 answer:
MrRissso [65]3 years ago
8 0

Answer:

a) t = 0.0185 s = 18.5 ms

b) T = 874.8 N

Explanation:

a)

First we find the seed of wave:

v = fλ

where,

v = speed of wave

f = frequency = 810 Hz

λ = wavelength = 0.4 m

Therefore,

v = (810 Hz)(0.4 m)

v = 324 m/s

Now,

v = L/t

where,

L = length of wire = 6 m

t = time taken by wave to travel length of wire

Therefore,

324 m/s = 6 m/t

t = (6 m)/(324 m/s)

<u>t = 0.0185 s = 18.5 ms</u>

<u></u>

b)

From the formula of fundamental frquency, we know that:

Fundamental Frequency = v/2L = (1/2L)(√T/μ)

v = √(T/μ)

where,

T = tension in string

μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹

Therefore,

324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)

(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹

<u>T = 874.8 N</u>

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Answer:

D)     D = \frac{5}{4}  - \frac{3}{4} \ C, E)  (C, D) = ( \frac{17}{7}, \ \frac{-4}{7}

Explanation:

Part D) two expressions are indicated

          3C + 4D = 5

          2C +5 D = 2

let's simplify each expression

         3C + 4D = 5

         4D = 5 - 3C

we divide by 4

            D = \frac{5}{4}  - \frac{3}{4} \ C

The other expression

       2C +5 D = 2

       2C = 2 - 5D

        C = 1 -  \frac{5}{2} \ D

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

          D =  \frac{5}{4}  - \frac{3}{4} \ C

we substitute in the other equation

            2C +5 D = 2

             2C +5 (5/4 - ¾ C) = 2

we solve

            C (2 - 15/4) + 25/4 = 2

             -7 / 4 C = 2 - 25/4

             -7 / 4 C = -17/4

              7C = 17

               C = \frac{17}{7}

now we calculate D

               D = \frac{5}{4} - \frac{3}{4} \ \frac{17}{7}

               D = 5/4 - 51/28

               D =\frac{35-51}{28}

               D = - 16/28

               D = - \frac{4}{7}

the result is (C, D) = ( \frac{17}{7}, \ \frac{-4}{7} )

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Transverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What i
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Answer:

0.055 kg

Explanation:

Given that

Length of the string, l = 5 m

Speed of the wave, v = 30 m/s

Tension on the string, F(t) = 10N

From the formula written in the attachment, we have

v = velocity of the wave, in m/s

F(t) = Tension on the string, in N

U = Mass per length of the string, in kg/m

m = Mass of the string, in kg

l = Length of the string, in m

See attachment for the calculation

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