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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. Distance between the two points having amplitude 2 mm is:
- a)1 m
- b)75cm
- c)60cm
- d)50 cm
Correct 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 fun...
Equation of standing wave ( As x=0 is taken as a node)
y = 2A sin kx cos ωt
y= A as amplitude is 2A
A = 2A sin kx
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A string of length 1.5 m with its two ends clamped is vibrating in fun...
**Given:**
- Length of the string = 1.5 m
- Amplitude at the center of the string = 4 mm
- Amplitude at two points = 2 mm
**To find:**
- Distance between the two points having an amplitude of 2 mm
**Solution:**
The fundamental frequency of a string fixed at both ends is given by the formula:
f = (v/2L)
Where,
- f is the frequency of vibration
- v is the velocity of the wave
- L is the length of the string
In the fundamental mode, the wavelength (λ) is double the length of the string. So,
λ = 2L
The velocity of the wave is given by the formula:
v = fλ
Substituting the value of λ, we get:
v = f * 2L
In the fundamental mode, the amplitude is maximum at the center of the string and zero at the ends. So, the amplitude decreases linearly from the center to the ends.
Let's consider the distance (x) from the center of the string to a point where the amplitude is 2 mm.
Using the amplitude formula,
A = (2mm) = (4mm) * (1 - 2x/L)
Simplifying the equation,
1/2 = 1 - 2x/L
2x/L = 1/2
x/L = 1/4
So, the distance between the two points having an amplitude of 2 mm is 1/4 times the length of the string.
Substituting the given length of the string, we get:
x = (1/4) * 1.5m
x = 0.375m
Converting to centimeters,
x = 0.375m * 100cm/m
x = 37.5cm
Therefore, the distance between the two points having an amplitude of 2 mm is 37.5 cm, which is closest to option 'A' (1m).
- Length of the string = 1.5 m
- Amplitude at the center of the string = 4 mm
- Amplitude at two points = 2 mm
**To find:**
- Distance between the two points having an amplitude of 2 mm
**Solution:**
The fundamental frequency of a string fixed at both ends is given by the formula:
f = (v/2L)
Where,
- f is the frequency of vibration
- v is the velocity of the wave
- L is the length of the string
In the fundamental mode, the wavelength (λ) is double the length of the string. So,
λ = 2L
The velocity of the wave is given by the formula:
v = fλ
Substituting the value of λ, we get:
v = f * 2L
In the fundamental mode, the amplitude is maximum at the center of the string and zero at the ends. So, the amplitude decreases linearly from the center to the ends.
Let's consider the distance (x) from the center of the string to a point where the amplitude is 2 mm.
Using the amplitude formula,
A = (2mm) = (4mm) * (1 - 2x/L)
Simplifying the equation,
1/2 = 1 - 2x/L
2x/L = 1/2
x/L = 1/4
So, the distance between the two points having an amplitude of 2 mm is 1/4 times the length of the string.
Substituting the given length of the string, we get:
x = (1/4) * 1.5m
x = 0.375m
Converting to centimeters,
x = 0.375m * 100cm/m
x = 37.5cm
Therefore, the distance between the two points having an amplitude of 2 mm is 37.5 cm, which is closest to option 'A' (1m).
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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. Distance between the two points havingamplitude 2 mm is:a)1 mb)75cmc)60cmd)50 cmCorrect answer is option 'A'. Can you explain this answer?
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