Two parts of a sonometer wire divided by a movable bridge differ in length by 0.2 cm and produce one beat per second, when sounded together. The total length of wire is 1 m, then the frequencies are

250.5 and 249.5 Hz

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230.5 and 229.5 Hz

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220.5 and 219.5 Hz

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210.5 and 209.5 Hz

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Solution

The correct option is C 250.5 and 249.5 Hz
Frequency, $f = \frac{1}{2 l} = \frac{1}{2 \times 0.2 \times 10^{- 2}} = 250 H z$
As beats are produced at one beat/sec:
$\Rightarrow$ Frequencies, $n_{A} = 250.5$ and $n_{B} = 249.5$
Such that: $n_{A} - n_{B} = 1 H z$

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Two parts of a sonometer wire divided by a movable bridge differ in length by 0.2 cm and produce one beat per second, when sounded together. The total length of wire is 1 m, then the frequencies are

The correct option is C 250.5 and 249.5 HzFrequency, f=12l=12×0.2×10−2=250 HzAs beats are produced at one beat/sec:⇒ Frequencies, nA=250.5 and nB=249.5Such that: nA−nB=1 Hz

The correct option is C 250.5 and 249.5 Hz
Frequency, $f = \frac{1}{2 l} = \frac{1}{2 \times 0.2 \times 10^{- 2}} = 250 H z$
As beats are produced at one beat/sec:
$\Rightarrow$ Frequencies, $n_{A} = 250.5$ and $n_{B} = 249.5$
Such that: $n_{A} - n_{B} = 1 H z$