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A wire 50 cm long is stretched by 60N . The mass per unit length of the wire is 1×10^-5 kg/m . Calculate the frequency of fundamental mode emitted by vibrating string.?
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A wire 50 cm long is stretched by 60N . The mass per unit length of th...
The fundamental frequency of the first node in a standing wave in a wire is given by,

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A wire 50 cm long is stretched by 60N . The mass per unit length of th...

Calculating Frequency of Fundamental Mode in a Vibrating String

Calculating Tension in the Wire:
- Given length of the wire, L = 50 cm = 0.5 m
- Tension in the wire, T = 60 N

Calculating Mass of the Wire:
- Mass per unit length, μ = 1×10^-5 kg/m
- Mass of the wire, m = μL = 1×10^-5 kg/m * 0.5 m = 5×10^-6 kg

Calculating Speed of Wave in the Wire:
- Speed of wave, v = √(T/μ) = √(60 N / 5×10^-6 kg) = √(1.2×10^7 m/s) = 3454.1 m/s

Calculating Frequency of Fundamental Mode:
- Fundamental frequency, f = v / 2L = 3454.1 m/s / (2*0.5 m) = 3454.1 Hz

Therefore, the frequency of the fundamental mode emitted by the vibrating string is 3454.1 Hz.
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A wire 50 cm long is stretched by 60N . The mass per unit length of the wire is 1×10^-5 kg/m . Calculate the frequency of fundamental mode emitted by vibrating string.?
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