Home
/
Physics
/
2. The speed of a wave travelling along a 0.65 m guitar string is 206m/s At n=6 the frequency is 950 Hz. (a) What is the frequency if a string with a wave speed of 150m/s is used? [ans: 690 Hz] (b) What is the frequency if the string is tightened to make the wave speed 350m/s [ans: 1600 Hz]

Question

2. The speed of a wave travelling along a 0.65 m guitar string is 206m/s At n=6 the frequency is 950 Hz. (a) What is the frequency if a string with a wave speed of 150m/s is used? [ans: 690 Hz] (b) What is the frequency if the string is tightened to make the wave speed 350m/s [ans: 1600 Hz]

2. The speed of a wave travelling along a 0.65 m guitar string is 206m/s At n=6 the
frequency is 950 Hz.
(a) What is the frequency if a string with a wave speed of 150m/s is used? [ans: 690 Hz]
(b) What is the frequency if the string is tightened to make the wave speed
350m/s [ans: 1600 Hz]

Solution

expert verifiedExpert Verified
4.4(180 Voting)
avatar
LeonaElite · Tutor for 8 years

Answer

Here's how to solve this problem:<br /><br />**Understanding the Relationship**<br /><br />The fundamental frequency (n=1) of a vibrating string is related to its wave speed (v) and length (L) by the following equation:<br /><br />f₁ = v / (2L)<br /><br />For higher harmonics (n=2, 3, 4...), the frequency (fₙ) is simply a multiple of the fundamental frequency:<br /><br />fₙ = n * f₁ = n * v / (2L)<br /><br />**Solving the Problem**<br /><br />**(a) Finding the frequency with a wave speed of 150 m/s**<br /><br />1. **Find the fundamental frequency (f₁) for the original string:** We know f₆ = 950 Hz and n=6. So, f₁ = f₆ / 6 = 950 Hz / 6 ≈ 158.33 Hz<br /><br />2. **Find the length (L) of the string:** We can rearrange the fundamental frequency equation to solve for L: L = v / (2f₁) = 206 m/s / (2 * 158.33 Hz) ≈ 0.65 m (This confirms the given length).<br /><br />3. **Calculate the new fundamental frequency (f₁') with the new wave speed (v' = 150 m/s):** f₁' = v' / (2L) = 150 m/s / (2 * 0.65 m) ≈ 115.38 Hz<br /><br />4. **Calculate the new frequency at n=6 (f₆'):** f₆' = n * f₁' = 6 * 115.38 Hz ≈ 692.3 Hz (This is approximately 690 Hz as requested by the question).<br /><br /><br />**(b) Finding the frequency with a wave speed of 350 m/s**<br /><br />We'll follow a similar process:<br /><br />1. **We already know the fundamental frequency (f₁) and length (L) from part (a).**<br /><br />2. **Calculate the new fundamental frequency (f₁'') with the new wave speed (v'' = 350 m/s):** f₁'' = v'' / (2L) = 350 m/s / (2 * 0.65 m) ≈ 269.23 Hz<br /><br />3. **Calculate the new frequency at n=6 (f₆''):** f₆'' = n * f₁'' = 6 * 269.23 Hz ≈ 1615.38 Hz (This is approximately 1600 Hz as requested by the question).<br /><br /><br />Therefore, the answers are approximately **690 Hz** and **1600 Hz**. The slight differences are due to rounding in intermediate steps.<br />
Click to rate: