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Standard XII

Physics

Beat Frequency

Two tuning fo...

Question

Two tuning forks when sounded together produce $6$ beats/s. The first fork has the frequency $3$ % higher than a standard one and the second has the frequency $2$ % less than the standard fork. The frequencies for the forks are

126.3, 120.3 H z

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162.7, 156.7 H z

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136.2, 130.2 H z

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123.6, 117.6 H z

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Solution

The correct option is D $123.6, 117.6 H z$
$(f + \frac{3 f}{100}) - (f - \frac{2 f}{100}) = 6$
$\Rightarrow 5 f = 600$
or
$f = 120 H z$
$f + \frac{3 f}{100} = 123.6$ Hz,
$f - \frac{2 f}{100} = 120 - \frac{2 \times 120}{100} = 117.6$ Hz,

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Two tuning forks when sounded together produce 6 beats/s. The first fork has the frequency 3% higher than a standard one and the second has the frequency 2% less than the standard fork. The frequencies for the forks are

The correct option is D 123.6,117.6Hz(f+3f100)−(f−2f100)=6⇒5f=600 orf=120Hzf+3f100=123.6Hz,f−2f100=120−2×120100=117.6Hz,

Two tuning forks when sounded together produce $6$ beats/s. The first fork has the frequency $3$ % higher than a standard one and the second has the frequency $2$ % less than the standard fork. The frequencies for the forks are

The correct option is D $123.6, 117.6 H z$
$(f + \frac{3 f}{100}) - (f - \frac{2 f}{100}) = 6$
$\Rightarrow 5 f = 600$
or
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$f + \frac{3 f}{100} = 123.6$ Hz,
$f - \frac{2 f}{100} = 120 - \frac{2 \times 120}{100} = 117.6$ Hz,