A sinometer wire of 1m length between the fixed ends. Find distances between bridges to be placed below the wire so that the there segments of wire have thier fundamental frequencies in the ratio 1:2:3?

To determine the distances between bridges that will create segments of a 1-meter sinometer wire with fundamental frequencies in the ratio of 1:2:3, we need to understand how the frequency of a vibrating string relates to its length. The fundamental frequency of a vibrating wire is inversely proportional to its length; that is, shorter segments vibrate at higher frequencies. Let's break this down step by step.

Understanding Fundamental Frequency

The fundamental frequency (f) of a vibrating wire can be expressed as:

f = (1/2L) * √(T/μ)

Where:

• L = length of the wire segment
• T = tension in the wire
• μ = linear mass density of the wire

For our purposes, we can assume that tension and linear mass density remain constant across the segments. Thus, the frequency is primarily dependent on the length of each segment.

Setting Up the Problem

Let’s denote the lengths of the three segments as L1, L2, and L3. According to the problem, we want the frequencies to be in the ratio 1:2:3. This means:

f1 : f2 : f3 = 1 : 2 : 3

Since frequency is inversely proportional to length, we can express this as:

L1 : L2 : L3 = 3 : 1.5 : 1

Calculating Segment Lengths

Let’s introduce a variable k to represent a common factor for the lengths:

L1 = 3k

L2 = 1.5k

L3 = k

Now, we know that the total length of the wire is 1 meter:

L1 + L2 + L3 = 1

Substituting the expressions for L1, L2, and L3 gives:

3k + 1.5k + k = 1

This simplifies to:

5.5k = 1

From this, we can solve for k:

k = 1/5.5 = 2/11

Finding Each Segment Length

Now we can find the lengths of each segment:

• L1 = 3k = 3 * (2/11) = 6/11 m
• L2 = 1.5k = 1.5 * (2/11) = 3/11 m
• L3 = k = 2/11 m

Determining the Distances Between Bridges

To find the distances between the bridges, we need to calculate the positions where the bridges will be placed along the 1-meter wire:

• The first bridge will be at L1 = 6/11 m from the left end.
• The second bridge will be at L1 + L2 = 6/11 + 3/11 = 9/11 m from the left end.
• The third bridge will be at the end of the wire, which is at 1 m.

Final Distances

Thus, the distances between the bridges are:

• First bridge: 6/11 m
• Second bridge: 9/11 m
• Third bridge: 1 m

In summary, by segmenting the wire into lengths of 6/11 m, 3/11 m, and 2/11 m, we achieve the desired frequency ratios of 1:2:3. This method illustrates the relationship between length and frequency in a straightforward manner, allowing for practical applications in musical instruments and other fields involving vibrating strings.