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A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:
- a)1 m
- b)75 cm
- c)60 cm
- d)50 cm
Correct answer is option 'A'. Can you explain this answer?
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Solution:
Given, length of the string, L = 1.5 m
Amplitude at the centre, A = 4 mm = 0.004 m
Amplitude at some point x from the centre, a = 2 mm = 0.002 m
We know that in a string vibrating in fundamental mode, the displacement y(x,t) can be given by the equation,
y(x,t) = A sin(kx) sin(ωt)
where, k = π/L and ω = πv/L
Here, v is the speed of the wave.
Let's find the value of v first.
We know that the wave speed, v = fλ
where, f is the frequency and λ is the wavelength.
In fundamental mode, the wavelength is twice the length of the string, i.e., λ = 2L.
Also, the frequency, f = v/λ = v/2L
Hence, ω = πv/L = 2πf
Now, let's find the value of k for the given string.
k = π/L = π/1.5 = 2.094 rad/m
Now, let's find the two points on the string where the amplitude is equal to 2 mm.
Let the distance of the first point from the centre be x1.
Then, a = A sin(kx1)
0.002 = 0.004 sin(2.094x1)
x1 = 0.2969 m
Similarly, let the distance of the second point from the centre be x2.
Then, a = A sin(kx2)
0.002 = 0.004 sin(2.094x2)
x2 = 1.2031 m
The minimum distance between the two points is given by,
d = x2 - x1
d = 1.2031 - 0.2969 = 0.9062 m
Therefore, the minimum distance between the two points having amplitude 2 mm is 0.9062 m, which is closest to option A (1 m).
Given, length of the string, L = 1.5 m
Amplitude at the centre, A = 4 mm = 0.004 m
Amplitude at some point x from the centre, a = 2 mm = 0.002 m
We know that in a string vibrating in fundamental mode, the displacement y(x,t) can be given by the equation,
y(x,t) = A sin(kx) sin(ωt)
where, k = π/L and ω = πv/L
Here, v is the speed of the wave.
Let's find the value of v first.
We know that the wave speed, v = fλ
where, f is the frequency and λ is the wavelength.
In fundamental mode, the wavelength is twice the length of the string, i.e., λ = 2L.
Also, the frequency, f = v/λ = v/2L
Hence, ω = πv/L = 2πf
Now, let's find the value of k for the given string.
k = π/L = π/1.5 = 2.094 rad/m
Now, let's find the two points on the string where the amplitude is equal to 2 mm.
Let the distance of the first point from the centre be x1.
Then, a = A sin(kx1)
0.002 = 0.004 sin(2.094x1)
x1 = 0.2969 m
Similarly, let the distance of the second point from the centre be x2.
Then, a = A sin(kx2)
0.002 = 0.004 sin(2.094x2)
x2 = 1.2031 m
The minimum distance between the two points is given by,
d = x2 - x1
d = 1.2031 - 0.2969 = 0.9062 m
Therefore, the minimum distance between the two points having amplitude 2 mm is 0.9062 m, which is closest to option A (1 m).
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A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer?
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A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer?.
A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer?.
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A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:a)1 m b)75 cmc)60 cmd)50 cmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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