1. A 0.44 m length of rope has one fixed end and one free end A wave moves along the rope at the speed 350m/s with a frequency of 200.0 Hz at n=1 . Tri (a) What is L_(1) if the frequency is doubled? [ans: L_(1)=0.22m] (b) What is the length of the string if n=3 [ans: 2.2 m] (c) What is L_(1) if the speed of the wave on the string is reduced to 200m/s ? [ans: 0.25 m]

**(a) Finding L₁ when frequency is doubled:**<br /><br />* **Understanding the relationship:** The fundamental frequency (n=1) of a wave on a string with one fixed end is given by f = v / (4L), where f is the frequency, v is the wave speed, and L is the length of the string. When the frequency is doubled (and the speed remains constant), the length must be halved to maintain the relationship.<br /><br />* **Calculation:** Since the original length is 0.44 m, the new length L₁ when the frequency is doubled is: L₁ = 0.44 m / 2 = 0.22 m<br /><br />* **Answer:** L₁ = 0.22 m<br /><br />**(b) Finding the length when n=3:**<br /><br />* **Understanding the relationship:** For higher harmonics (n=2, 3, 4...), the formula becomes f = (nv) / (4L). We are given n=3.<br /><br />* **Calculation:** We can rearrange the formula to solve for L: L = (nv) / (4f). Plugging in the given values (n=3, v=350 m/s, f=200 Hz): L = (3 * 350 m/s) / (4 * 200 Hz) = 1.3125 m. However, this length is for the *fundamental* frequency associated with the n=3 harmonic. The question asks for the *total* length of the string when n=3. The total length for the nth harmonic is n times the fundamental length for that harmonic. In this case, the fundamental length for n=3 is 0.44m (as given in the problem). Therefore, for n=3, the total length is 3 * 0.44m = 1.32m. There seems to be a slight discrepancy due to rounding in the intermediate calculations. Using the unrounded value of 1.3125m and multiplying by 3 gives 3.9375m. If we use the given frequency and speed to calculate the fundamental wavelength, we get λ = v/f = 350/200 = 1.75m. For n=3, the length of the string is 3λ/4 = 3(1.75)/4 = 1.3125m. So the length of the string when n=3 is 1.3125m, not 2.2m as stated in the question.<br /><br />* **Answer:** 1.31 m (The answer provided in the question is incorrect.)<br /><br />**(c) Finding L₁ when the speed is reduced:**<br /><br />* **Understanding the relationship:** Returning to the fundamental frequency formula, f = v / (4L), we see that if the speed (v) decreases, the length (L) must also decrease to maintain the same frequency.<br /><br />* **Calculation:** Let's denote the initial speed and length as v₁ and L₁, and the final speed and length as v₂ and L₂. We have the relationship: v₁ / L₁ = v₂ / L₂. Plugging in the values (v₁ = 350 m/s, L₁ = 0.44 m, v₂ = 200 m/s): (350 m/s) / (0.44 m) = (200 m/s) / L₂. Solving for L₂: L₂ = (200 m/s * 0.44 m) / (350 m/s) ≈ 0.251 m<br /><br />* **Answer:** L₁ ≈ 0.25 m<br />