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A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.
Correct answer is '5'. Can you explain this answer?
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A 20 cm long string, having a mass of 1.0 g, is fixed at both the end...
Given:
- Length of the string (L) = 20 cm
- Mass of the string (m) = 1.0 g = 0.001 kg
- Tension in the string (T) = 0.5 N
- Frequency of the vibrator (f) = 100 Hz
To Find:
The separation between the successive nodes on the string.
Solution:
1. Speed of the Wave:
The speed of the wave on the string can be calculated using the formula:
v = √(T/μ)
where v is the speed of the wave and μ is the linear mass density of the string.
The linear mass density of the string can be calculated using the formula:
μ = m/L
Substituting the given values, we get:
μ = 0.001 kg / 0.2 m = 0.005 kg/m
Now, calculating the speed of the wave:
v = √(0.5 N / 0.005 kg/m) = √(100 m^2/s^2) = 10 m/s
2. Wavelength of the Wave:
The wavelength (λ) of the wave can be calculated using the formula:
λ = v/f
where λ is the wavelength, v is the speed of the wave, and f is the frequency of the vibrator.
Substituting the given values, we get:
λ = 10 m/s / 100 Hz = 0.1 m = 10 cm
3. Node Separation:
The separation between the successive nodes on the string is half the wavelength (λ/2).
Substituting the value of λ, we get:
Node Separation = 10 cm / 2 = 5 cm
Answer:
The separation between the successive nodes on the string is 5 cm.
- Length of the string (L) = 20 cm
- Mass of the string (m) = 1.0 g = 0.001 kg
- Tension in the string (T) = 0.5 N
- Frequency of the vibrator (f) = 100 Hz
To Find:
The separation between the successive nodes on the string.
Solution:
1. Speed of the Wave:
The speed of the wave on the string can be calculated using the formula:
v = √(T/μ)
where v is the speed of the wave and μ is the linear mass density of the string.
The linear mass density of the string can be calculated using the formula:
μ = m/L
Substituting the given values, we get:
μ = 0.001 kg / 0.2 m = 0.005 kg/m
Now, calculating the speed of the wave:
v = √(0.5 N / 0.005 kg/m) = √(100 m^2/s^2) = 10 m/s
2. Wavelength of the Wave:
The wavelength (λ) of the wave can be calculated using the formula:
λ = v/f
where λ is the wavelength, v is the speed of the wave, and f is the frequency of the vibrator.
Substituting the given values, we get:
λ = 10 m/s / 100 Hz = 0.1 m = 10 cm
3. Node Separation:
The separation between the successive nodes on the string is half the wavelength (λ/2).
Substituting the value of λ, we get:
Node Separation = 10 cm / 2 = 5 cm
Answer:
The separation between the successive nodes on the string is 5 cm.
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Community Answer
A 20 cm long string, having a mass of 1.0 g, is fixed at both the end...
Velocity of sound in the string = v = 
; where T is the tension and μ is the mass per unit length.
; where T is the tension and μ is the mass per unit length.
Distance between successive nodes,
λ/2 = 5 cm
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A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. Can you explain this answer?
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A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. 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 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. 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 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. Can you explain this answer?.
A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. 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 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. 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 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. Can you explain this answer?.
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A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. Can you explain this answer?, a detailed solution for A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. Can you explain this answer? has been provided alongside types of A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.Correct answer is '5'. Can you explain this answer? theory, EduRev gives you an
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