Standing wave is produced in 10m long stretched string. If the string vibrates in 5 segments and wave velocity is 20m/s, then frequency of oscillation is
A
5Hz
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B
2Hz
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C
8Hz
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D
7Hz
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Solution
The correct option is A5Hz Given, string is vibrating in 5 segments.
We know that length of 1 loop or segment is equal to λ2. ⇒L=5×(λ2) ⇒10=5λ2 ⇒λ=4m
So, f=vλ=204=5Hz