Physics, asked by Anonymous, 10 months ago

Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 sec and the velocity of the wave is 300 m/sec. What is the phase difference between the oscillations of two points?

Answers

Answered by Anonymous
87

Question :

Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillation of two points?

Theory :

pase diffrence :

At any instant t, if Φ1 and Φ2 are the phases of two particles with distances from the origin x1 and x2 . Then ,

Φ1 = (wt-kx1)

and

Φ2 = (wt-kx2)

⇒ Φ1- Φ2 = k(x2- x1)

{\purple{\boxed{\large{\bold{</p><p>Phase\: Difference = \frac{2\pi}{ \lambda}  \times path \: diffrence \:}}}}}

 \huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}

Given:

Time period ,T = 0.05 sec

velocity of wave ,v = 300 m/s

x _{2} = 15 m

x _{1} = 10m

frequency , f = \frac{1}{T}

f =  \frac{1}{0.05}

f =  \frac{100}{5}

⇒f = 20Hz

Δ x = 15-10 = 5 m

we know that

λ = v/f

 λ=  \frac{300}{20}

\lambda  = 15 m

Phase\: Difference = \frac{2\pi}{ \lambda}  \times path \: diffrence \:

put the value ofλ and Δx

phase \: diffrence \:  =  \frac{2\pi}{ \lambda}  \times path \: diffrence \:

 =  \frac{2\pi}{15}  \times 5

 =  \frac{2\pi}{3}

{\purple{\boxed{\large{\bold{Phase\: Difference = \frac{2\pi}{ 3}  </p><p>}}}}}

Answered by Avni2348
4

Answer :

The phase difference between the oscillations of two points is 2.093

Explanation:

Given path difference = 15 m - 10 m = 5 m

The time period of oscillation T= 0.05 s.

Frequency,

Velocity of wave v= 300 m/s

So wavelength,

To calculate the phase difference use relation between path and phase difference,

Phase difference   × path difference

Substitute the given values, we get

Phase difference

Phase difference

Thus, phase difference is 2.093.

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