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Multiplication & Division of Errors

In resonance ...

Question

In resonance tube experiment, the velocity of sound is given by $v = 2 f_{0} (l_{1} - l_{1}) .$ We found $l_{1} = 25.0 c m$ and $l_{2} = 75.0 c m$ . If there is no error in frequency, what will be the maximum permissible error in the speed of sound? $(T a k e f_{0} = 325 H z)$

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Solution

$V = 2 f_{0} (l_{2} - l_{1})$
$(d V) = 2 f_{0} (d t_{2} - d t_{1})$
$(d V)_{m a x} = m a x o f [2 f_{0} (\pm △ l_{2} \pm △ l_{2})]$
= $2 f_{0} (△ l_{2} + △ l_{1})$
$l_{1} = 25.0 c m \Rightarrow l_{1} = 0.1 c m$ (place value of last number)
$l_{1} = 75.0 c m \Rightarrow l_{2} = 0.1 c m$ (place value of last number)
So the maximum permissible error in the speed of sound $(d V)_{m a x} = 2 (325 H z) (0.1 c m + 0.1 c m) = 1.3 m s^{- 1}$
Value of V = $2 f_{0} (l_{2} - l_{1}) = 2 (325 H z) (75.0 c m - 25.0 c m) = 325 m s^{- 1}$
So $V = (325 \pm 1.3) m s^{- 1}$ .

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Similar questions

l_{1} = 25.0

l_{2} = 75.0

f_{0} = 325 H z

V = 2 f_{0} (l_{2} - l_{1})

v = 2 n (l_{2} - l_{1})

l_{1}, l_{2}

l_{1} = 25.0 c m, l_{2} = 75.0 c m

l_{1} = 25 cm and l_{2} = 75 cm

l

0.1 cm

f_{0} = 325 Hz

512 H z

30.7 c m

63.2 c m

332 m / s

l_{1} = 30.7 c m

l_{2} = 63.2 c m

f = 512 H z

v = f λ

λ = 2 (l_{2} - l_{1})

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