A stationary wave is set up on a stretched string of length $l$ and diameter $d$ - AQA - A-Level Physics - Question 15 - 2017 - Paper 1

To find the length and diameter of the second string required for it to have the same first-harmonic frequency as the first string, we start with the formula for the frequency of a vibrating string:

f = rac{1}{2L} imes ext{v}

where:

• $f$ is the frequency,
• $L$ is the length of the string,
• $v$ is the wave speed on the string, given by:

ho} $$ Here $T$ is the tension in the string, and $ ho$ is the linear mass density, which can be expressed as:

ho = rac{m}{L} = rac{ rac{ ext{mass}}{V}}{L} = rac{ ext{density} imes ext{volume}}{L} = rac{ ext{density} imes (rac{ ext{pi}d^2}{4} imes L)}{L} = rac{ ext{density} imes ext{pi}}{4} d^2 $$

Since both strings have the same tension, the frequency of the two strings can be equated for the same first-harmonic: rac{1}{2l} imes v_1 = rac{1}{2L_2} imes v_2

Given that the wave speed method depends on the diameter, we also find the relation between diameter and first-harmonic:

determining ratio of lengths: rac{L_2}{l} = rac{d^2}{d_2^2}

By simplifying and equating: L_2 = l imes rac{d_2^2}{d^2}

The values can then be matched with the options, leading to:

• If the first string diameter is $d$ and we want the second to attain the same length and diameter requirements, from options, we can conclude the second string would have to be of length rac{l}{2} and diameter $2d$.