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Standard XII

Physics

Ropes

Two strings ‘...

Question

Two strings ‘x’ and ‘y’ on a Veena playing the same note are slightly out of tune and produce 6 beats per second. The tension in ‘x’ is slightly decreased and it is found that the beats fall to 3 per second. If the original frequency of ‘x’ is 324 Hz, the frequency of ‘y’ is

318 Hz

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330 Hz

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518 Hz

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618 Hz

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Solution

The correct option is A 318 Hz
$f_{x} = 324$
As tension in $x$ is decreased frequency of $x$ decreases and as beats fall with decrease in frequency of $x$ .
So, frequency of $x$ is greater than frequency of $y$ .
$B e a t s = δ_{x} - δ_{y}$
$6 = 324 - δ_{y}$
$\Rightarrow δ_{y} = 318 H_{z}$

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Two strings ‘x’ and ‘y’ on a Veena playing the same note are slightly out of tune and produce 6 beats per second. The tension in ‘x’ is slightly decreased and it is found that the beats fall to 3 per second. If the original frequency of ‘x’ is 324 Hz, the frequency of ‘y’ is

The correct option is A 318 Hzfx=324As tension in x is decreased frequency of x decreases and as beats fall with decrease in frequency of x.So, frequency of x is greater than frequency of y.Beats=δx−δy6=324−δy⇒δy=318 Hz

The correct option is A 318 Hz
$f_{x} = 324$
As tension in $x$ is decreased frequency of $x$ decreases and as beats fall with decrease in frequency of $x$ .
So, frequency of $x$ is greater than frequency of $y$ .
$B e a t s = δ_{x} - δ_{y}$
$6 = 324 - δ_{y}$
$\Rightarrow δ_{y} = 318 H_{z}$