Photo AI
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
Question 15
A stationary wave is set up on a stretched string of length $l$ and diameter $d$. Another stationary wave is also set up on a second string made from the same materi... show full transcript
Worked Solution & Example Answer: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
Step 1
What length and diameter are required for the second string so that both strings have the same first-harmonic frequency?
Answer
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:
- is the frequency,
- is the length of the string,
- is the wave speed on the string, given by:
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 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 .