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A particular guitar wire is 30.0 cm long...

A particular guitar wire is 30.0 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes ?

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To solve the problem, we need to determine how far from the end of the guitar string the finger must be placed to play the higher notes. We will use the relationship between frequency and the length of the vibrating string. ### Step-by-Step Solution: 1. **Identify Given Values:** - Length of the guitar wire (L1) = 30.0 cm = 0.3 m - Frequency when no finger is placed (f1) = 196 Hz - Higher frequencies (f2, f3, f4, f5) = 220 Hz, 247 Hz, 262 Hz, and 294 Hz 2. **Understand the Relationship:** - The frequency of a vibrating string is inversely proportional to its length. This can be expressed as: \[ \frac{f_1}{f_n} = \frac{L_n}{L_1} \] - Where \(L_n\) is the new length of the string when the finger is placed to play the note corresponding to frequency \(f_n\). 3. **Calculate New Lengths:** - For each frequency, we can rearrange the formula to find \(L_n\): \[ L_n = L_1 \cdot \frac{f_1}{f_n} \] 4. **Calculate for Each Frequency:** - **For f2 = 220 Hz:** \[ L_2 = 0.3 \cdot \frac{196}{220} = 0.267 \text{ m} = 26.7 \text{ cm} \] - **For f3 = 247 Hz:** \[ L_3 = 0.3 \cdot \frac{196}{247} = 0.238 \text{ m} = 23.8 \text{ cm} \] - **For f4 = 262 Hz:** \[ L_4 = 0.3 \cdot \frac{196}{262} = 0.224 \text{ m} = 22.4 \text{ cm} \] - **For f5 = 294 Hz:** \[ L_5 = 0.3 \cdot \frac{196}{294} = 0.20 \text{ m} = 20.0 \text{ cm} \] 5. **Determine Distance from the End of the String:** - The distance from the end of the string where the finger should be placed is given by: \[ d_n = L_1 - L_n \] - **For f2:** \[ d_2 = 0.3 - 0.267 = 0.033 \text{ m} = 3.3 \text{ cm} \] - **For f3:** \[ d_3 = 0.3 - 0.238 = 0.062 \text{ m} = 6.2 \text{ cm} \] - **For f4:** \[ d_4 = 0.3 - 0.224 = 0.076 \text{ m} = 7.6 \text{ cm} \] - **For f5:** \[ d_5 = 0.3 - 0.20 = 0.10 \text{ m} = 10.0 \text{ cm} \] ### Final Results: - To play the notes: - For 220 Hz: Place finger 3.3 cm from the end. - For 247 Hz: Place finger 6.2 cm from the end. - For 262 Hz: Place finger 7.6 cm from the end. - For 294 Hz: Place finger 10.0 cm from the end.

To solve the problem, we need to determine how far from the end of the guitar string the finger must be placed to play the higher notes. We will use the relationship between frequency and the length of the vibrating string. ### Step-by-Step Solution: 1. **Identify Given Values:** - Length of the guitar wire (L1) = 30.0 cm = 0.3 m - Frequency when no finger is placed (f1) = 196 Hz - Higher frequencies (f2, f3, f4, f5) = 220 Hz, 247 Hz, 262 Hz, and 294 Hz ...
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HC VERMA ENGLISH-WAVE MOTION AND WAVES ON A STRING-Exercises
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  2. The length of the wire shown in figure between the pulley is 1.5 m and...

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  4. A wire, fixed at both ends is seen to vibrate at a resonant frequency ...

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  5. A string, fixed at both ends, vibrates in a resonant mode with a separ...

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  7. A particular guitar wire is 30.0 cm long and vibrates at a frequency o...

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  8. A steel wire fixed at both ends has a fundamental frequency of 200 Hz....

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  9. Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Fin...

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  10. Two wires are kept tight between the same pair of supports. The tensio...

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  11. A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported ...

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  12. Figure shows an aluminium wire of length 60 cm joined to a steel wire ...

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  13. A string of length L fixed at both ends vibrates in its fundamental mo...

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  14. A 2 m-long string fixed at both ends is set into vibrations in its fir...

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  15. The equation for the vibration of a string, fixed at both ends vibrati...

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  16. The equation of a standing wave, produced on a string fixed at both en...

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  17. A 40 cm wire having a mass of 3.2 g is stretched between two fixed sup...

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  18. Figure shows a string stretched by a block going over a pulley. The st...

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  19. A 2.00 m-long rope, having a mass of 80 g, is fixed at one end and is ...

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