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A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.
- a)0cm
- b)1cm
- c)-1cm
- d)-2cm
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A sinusoidal wave has an amplitude of 2cm. It is travelling in the neg...
Π radians per centimeter.
The equation for a sinusoidal wave is given by:
y = A sin (ωt - kx + φ)
where:
- y is the displacement of the wave from its equilibrium position
- A is the amplitude of the wave
- ω is the angular frequency of the wave (in radians per second)
- t is the time
- k is the wave number (in radians per centimeter)
- x is the position of the wave along the x-axis
- φ is the phase angle
The distance between two crests is equal to the wavelength (λ). In this case, λ = 2cm. The wave number is given by k = 2π/λ = π radians per centimeter.
Since the wave is travelling in the negative x-direction, the phase angle is given by φ = π.
Therefore, the equation for the wave is:
y = 2 sin (πt - πx + π)
The angular frequency is given by ω = 2πf, where f is the frequency of the wave. The frequency is related to the wavelength by f = v/λ, where v is the speed of the wave. Since the wave is not moving, v = 0 and the frequency is also 0. Therefore, the angular frequency is 0 radians per second.
The equation for a sinusoidal wave is given by:
y = A sin (ωt - kx + φ)
where:
- y is the displacement of the wave from its equilibrium position
- A is the amplitude of the wave
- ω is the angular frequency of the wave (in radians per second)
- t is the time
- k is the wave number (in radians per centimeter)
- x is the position of the wave along the x-axis
- φ is the phase angle
The distance between two crests is equal to the wavelength (λ). In this case, λ = 2cm. The wave number is given by k = 2π/λ = π radians per centimeter.
Since the wave is travelling in the negative x-direction, the phase angle is given by φ = π.
Therefore, the equation for the wave is:
y = 2 sin (πt - πx + π)
The angular frequency is given by ω = 2πf, where f is the frequency of the wave. The frequency is related to the wavelength by f = v/λ, where v is the speed of the wave. Since the wave is not moving, v = 0 and the frequency is also 0. Therefore, the angular frequency is 0 radians per second.
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Community Answer
A sinusoidal wave has an amplitude of 2cm. It is travelling in the neg...
The equation suiting the given conditions will be: y = Asin(kx+wt+a)
At x=0, t=0, velocity is negative, so ‘a’ will be π.
k = 2π/λ = π.
∴ y = 2sin(πx + 2t + π).
At x=20cm & t=10s,
y = 2sin(20.5π + 10π + π)
= 2sin(31.5π)
= -2cm.
At x=0, t=0, velocity is negative, so ‘a’ will be π.
k = 2π/λ = π.
∴ y = 2sin(πx + 2t + π).
At x=20cm & t=10s,
y = 2sin(20.5π + 10π + π)
= 2sin(31.5π)
= -2cm.
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A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer?
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A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer?.
A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sinusoidal wave has an amplitude of 2cm. It is travelling in the negative x-direction. The distance between two crests is 2cm. The angular frequency is πs-1. What is the displacement of the particle at x = 20.5cm, at t = 10s? Assume that at t=0 & x=0 the particle was at the mean position and going downwards. ‘x’ &’y’ are in cm.a)0cmb)1cmc)-1cmd)-2cmCorrect answer is option 'D'. Can you explain this answer?.
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