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    Paper 1, Question 7.3

    I have no idea what they are doing in that one, so any help will be appreciated (and in as much simplicity as possible )
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    (Original post by 123chem)
    https://drive.google.com/folderview?...1k&usp=sharing

    Paper 1, Question 7.3

    I have no idea what they are doing in that one, so any help will be appreciated (and in as much simplicity as possible )
    Will walk you through it as a simpleish derivation:

    the relationship between the speed of a wave, it's frequency and wavelength is:

    \displaystyle f=\frac{v}{\lambda}

    from Newton's second law and some approximations (not going to walk you through this bit as it is complicated so just take the result as a given) the speed of a wave on a string is related to the tension,mass and length of a string by:

    \displaystyle v=\sqrt{\frac{TL}{m}}

    combining these gives us:

    \displaystyle f = \frac{1}{\lambda}\sqrt{\frac{TL}  {m}}

    We can rearrange this to get:

    \displaystyle T = (f\lambda)^2\frac{m}{L}

    we will now define  \lambda = 2l where l is the distance between the two fixed points (in this case X and Y). This relates the length between the points and the FUNDAMENTAL wavelength. Since we are using the fundamental wavelength, it also follows that we must use the fundamental frequency. (take it as a given that the tension remains the same regardless which mode we are in).

    We can see from the diagram that we are in mode 3. The distance XY is given as 0.66m. We can calculate the fundamental frequency as:

    \displaystyle f_{fund} = \frac{f_n}{n} = \frac{330Hz}{3} = 110Hz

    substituting in our relation between the fundamental wavelength and the distance XY into our earlier relation we also find that:

    \displaystyle T = (2fl)^2\frac{m}{L} = 4(fl)^2\frac{m}{L}

    we know the values of l = 0.66m, f = 110Hz, L = 0.91m and m = 3.1g = 3.1\mathrm{x}10^{-3}Kg and can plug these into our equation to get:

    \displaystyle T = 4 \mathrm{x} 110^2 \mathrm{x} 0.66^2 \mathrm{x} \frac {3.1\mathrm{x}10^{-3}}{0.91} = 71.8N
 
 
 
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