Question 310: Waves

Question

The length of a sonometer wire 'AB' is 110 cm . Where should the two bridges be placed from end ' $A$ ' to divide the wire in three segments whose fundamental frequencies are in the ratio $1: 2: 3$ ?

Accepted Answer

For a sonometer wire, the fundamental frequency ($f$) of vibration is inversely proportional to the length ($L$) of the vibrating segment. Thus, the frequency is given by:
$$ f \propto \frac{1}{L} $$
Given that the fundamental frequencies of the three segments are in the ratio $1: 2: 3$, we need to determine the lengths of these segments such that their inverses also follow the ratio $1: 2: 3$:
Let the lengths of the segments be $L_1$, $L_2$, and $L_3$. The ratio of their fundamental frequencies being $1:2:3$ implies:
$$ \frac{1}{L_1} : \frac{1}{L_2} : \frac{1}{L_3} = 1 : 2 : 3 $$
Inverting the terms in each ratio, we have:
$$ L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} $$
To simplify this ratio, find a common multiple of the denominators (2 and 3), which is 6. Therefore, multiplying each term by 6 gives:
$$ L_1 : L_2 : L_3 = 6 : 3 : 2 $$
Since the total length of the wire is 110 cm, we can write:
$$ L_1 + L_2 + L_3 = 110 $$
Substituting the ratio terms:
$$ 6x + 3x + 2x = 110 $$
Solving for $x$:
$$ 11x = 110 $$
$$ x = 10 $$
Thus, the lengths of the segments are:
$L_1 = 6x = 6 	imes 10 = 60$ cm
$L_2 = 3x = 3 	imes 10 = 30$ cm
$L_3 = 2x = 2 	imes 10 = 20$ cm
Therefore, the positions of the bridges from point $A$ should be at:
First bridge at 60 cm (end of the first segment)
Second bridge at 60 cm + 30 cm = 90 cm (end of the second segment)
Hence, the placement of the bridges should be at 60 cm and 90 cm.
Answer:
Option A
: 60 cm and 90 cm

Answer

60 cm and 90 cm

Answer

90 cm and 100 cm

Answer

40 cm and 80 cm

Answer

50 cm and 90 cm

Waves

The length of a sonometer wire 'AB' is 110 cm . Where should the two bridges be placed from end ' A ' to divide the wire in three segments whose fundamental frequencies are in the ratio 1: 2: 3 ?

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