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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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Given Information:
- Length of the string (L) = 1.5 m
- Amplitude at the center of the string = 4 mm
- Minimum amplitude required = 2 mm
Calculating Wavelength:
- In fundamental mode, the wavelength for a vibrating string is equal to twice the length of the string (λ = 2L).
- Therefore, wavelength (λ) = 2 * 1.5 = 3 m
Calculating Distance between Points:
- The distance between two points with the same amplitude on a vibrating string is equal to half the wavelength.
- Therefore, distance between two points with amplitude 2 mm = 0.5 * λ = 0.5 * 3 = 1.5 m
Converting Distance to Centimeters:
- 1.5 meters is equal to 150 centimeters.
- Therefore, the minimum distance between the two points having an amplitude of 2 mm is 150 cm.
Therefore, the correct answer is option 'A'
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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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