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A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?
Correct answer is '0.2'. Can you explain this answer?
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Verified Answer
A transverse harmonic wave of amplitude 0.02 m is generated at one end...
The general equation of this wave is given by
Here A = 0.02 m
Again, when x = 0.1m, y = -0.005 m
or
Again x = 0.2m, y = 0.005m
or λ = 0.2m
The correct answer is: 0.2
Here A = 0.02 m
Again, when x = 0.1m, y = -0.005 m
or
Again x = 0.2m, y = 0.005m
or λ = 0.2m
The correct answer is: 0.2
Most Upvoted Answer
A transverse harmonic wave of amplitude 0.02 m is generated at one end...
Given data:
Amplitude (A) = 0.02 m
Frequency (f) = 500 Hz
Position of particle at x = 0.1 m: displacement (y₁) = 0.005 m
Position of particle at x = 0.2 m: displacement (y₂) = 0.005 m
Understanding the problem:
We are given a transverse harmonic wave generated by a tuning fork. The wave is traveling along a horizontal string. We need to find the wavelength of the wave based on the given data.
Approach:
1. Calculate the wave number (k) using the formula: k = 2π/λ, where λ is the wavelength.
2. Determine the phase difference (Δφ) between the particles at x = 0.1 m and x = 0.2 m using the formula: Δφ = 2π(Δx)/λ, where Δx is the distance between the two particles.
3. Use the equation for a transverse harmonic wave: y = A sin(kx - ωt + φ), where ω = 2πf and φ is the phase constant.
4. Substitute the given values into the equation and solve for the phase constant (φ).
5. Use the calculated phase constant and the phase difference (Δφ) to find the value of k.
6. Finally, calculate the wavelength (λ) using the formula: λ = 2π/k.
Solution:
1. Calculate the wave number (k):
k = 2π/λ
2. Determine the phase difference (Δφ):
Δφ = 2π(Δx)/λ
3. Use the equation for a transverse harmonic wave:
y = A sin(kx - ωt + φ)
4. Substitute the given values into the equation:
y₁ = A sin(k(0.1) - ωt + φ)
y₂ = A sin(k(0.2) - ωt + φ)
5. Solve for the phase constant (φ) by subtracting the two equations:
y₁ - y₂ = A sin(k(0.1) - ωt + φ) - A sin(k(0.2) - ωt + φ)
0.005 - 0.005 = 0.02 sin(k(0.1) - ωt + φ) - 0.02 sin(k(0.2) - ωt + φ)
0 = 0.02 [sin(k(0.1) - ωt + φ) - sin(k(0.2) - ωt + φ)]
6. Since sin(A) - sin(B) = 2 cos[(A + B)/2] sin[(A - B)/2], we can rewrite the equation as:
0 = 0.04 cos[(k(0.1) - ωt + φ + k(0.2) - ωt + φ)/2] sin[(k(0.1) - ωt + φ - k(0.2) + ωt - φ)/2]
0 = 0.04 cos[(k(0.3)
Amplitude (A) = 0.02 m
Frequency (f) = 500 Hz
Position of particle at x = 0.1 m: displacement (y₁) = 0.005 m
Position of particle at x = 0.2 m: displacement (y₂) = 0.005 m
Understanding the problem:
We are given a transverse harmonic wave generated by a tuning fork. The wave is traveling along a horizontal string. We need to find the wavelength of the wave based on the given data.
Approach:
1. Calculate the wave number (k) using the formula: k = 2π/λ, where λ is the wavelength.
2. Determine the phase difference (Δφ) between the particles at x = 0.1 m and x = 0.2 m using the formula: Δφ = 2π(Δx)/λ, where Δx is the distance between the two particles.
3. Use the equation for a transverse harmonic wave: y = A sin(kx - ωt + φ), where ω = 2πf and φ is the phase constant.
4. Substitute the given values into the equation and solve for the phase constant (φ).
5. Use the calculated phase constant and the phase difference (Δφ) to find the value of k.
6. Finally, calculate the wavelength (λ) using the formula: λ = 2π/k.
Solution:
1. Calculate the wave number (k):
k = 2π/λ
2. Determine the phase difference (Δφ):
Δφ = 2π(Δx)/λ
3. Use the equation for a transverse harmonic wave:
y = A sin(kx - ωt + φ)
4. Substitute the given values into the equation:
y₁ = A sin(k(0.1) - ωt + φ)
y₂ = A sin(k(0.2) - ωt + φ)
5. Solve for the phase constant (φ) by subtracting the two equations:
y₁ - y₂ = A sin(k(0.1) - ωt + φ) - A sin(k(0.2) - ωt + φ)
0.005 - 0.005 = 0.02 sin(k(0.1) - ωt + φ) - 0.02 sin(k(0.2) - ωt + φ)
0 = 0.02 [sin(k(0.1) - ωt + φ) - sin(k(0.2) - ωt + φ)]
6. Since sin(A) - sin(B) = 2 cos[(A + B)/2] sin[(A - B)/2], we can rewrite the equation as:
0 = 0.04 cos[(k(0.1) - ωt + φ + k(0.2) - ωt + φ)/2] sin[(k(0.1) - ωt + φ - k(0.2) + ωt - φ)/2]
0 = 0.04 cos[(k(0.3)
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A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer?
Question Description
A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer?.
A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer?.
Solutions for A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer?, a detailed solution for A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? has been provided alongside types of A transverse harmonic wave of amplitude 0.02 m is generated at one end (x = 0) along horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of the particle at x = 0.1 m is 0.005 m and that of the particle at x = 0.2 m is 0.005 m. Calculate the wavelength of the wave?Correct answer is '0.2'. Can you explain this answer? theory, EduRev gives you an
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