A guitar string is 90 cm long and has a fundamental frequency of 124 Hz. Where should it be pressed to produce a fundamatal frequecy of 186 Hz?

Options

A. 60 cm
B. 30 cm
C. 20 cm
D. 10 cm

Explanation

To solve the problem step by step, we will use the relationship between the fundamental frequency of a vibrating string and its length. The fundamental frequency (f) is inversely proportional to the length (L) of the string, which can be expressed mathematically as:

\[ f_1 L_1 = f_2 L_2 \]

Where:
- \( f_1 \) is the initial frequency (124 Hz)
- \( L_1 \) is the initial length of the string (90 cm)
- \( f_2 \) is the new frequency (186 Hz)
- \( L_2 \) is the new length of the string we need to find.

### Step 1: Write down the known values
- \( L_1 = 90 \) cm
- \( f_1 = 124 \) Hz
- \( f_2 = 186 \) Hz

### Step 2: Set up the equation
Using the relationship \( f_1 L_1 = f_2 L_2 \), we can rearrange it to solve for \( L_2 \):
\[ L_2 = \frac{f_1 L_1}{f_2} \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation gives:
\[ L_2 = \frac{124 \, \text{Hz} \times 90 \, \text{cm}}{186 \, \text{Hz}} \]

### Step 4: Calculate \( L_2 \)
Now we perform the calculation:
1. Calculate the numerator:
\[ 124 \times 90 = 11160 \]
2. Divide by the denominator:
\[ L_2 = \frac{11160}{186} \]

Calculating this gives:
\[ L_2 \approx 60 \, \text{cm} \]

### Step 5: Conclusion
The string should be pressed at a point that shortens its effective length to 60 cm in order to produce a fundamental frequency of 186 Hz.

### Final Answer
The string should be pressed to produce a fundamental frequency of 186 Hz at a length of 60 cm.

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