In an experiment to study standing waves, you use a string whose mass per length is $μ$ = $(1.0 ± 0.1) × 10^{-4}\ kg/m$ . You look at the fundamental mode, whose frequency $f$ is related to the length $L$ and tension $T$ of the string by the following equation $L$ = $\frac{1}{{2f}}\sqrt {\frac{T}{\mu }} $ You make a plot with $L$ on the $y-$ axis and $\sqrt T$ on the $x-$ axis, and find that the best fitting line is $y$ = $(8.0 ± 0.3) × 10^{-3}x + (0.2 ± 0.04)$ in $SI\ units$ . What is the value of the frequency of the wave (including the error)? Express your result in $SI\ unit\ (Hz)$

Q: In an experiment to study standing waves, you use a string whose mass per length is $μ$ = $(1.0 ± 0.1) × 10^{-4}\ kg/m$ . You look at the fundamental mode, whose frequency $f$ is related to the length $L$ and tension $T$ of the string by the following equation $L$ = $\frac{1}{{2f}}\sqrt {\frac{T}{\mu }} $ You make a plot with $L$ on the $y-$ axis and $\sqrt T$ on the $x-$ axis, and find that the best fitting line is $y$ = $(8.0 ± 0.3) × 10^{-3}x + (0.2 ± 0.04)$ in $SI\ units$ . What is the value of the frequency of the wave (including the error)? Express your result in $SI\ unit\ (Hz)$

c $\frac{1}{2 \mathrm{f} \sqrt{\mu}}=8 \times 10^{-3}$ $\frac{1}{2 \times 10^{-2} \times 8 \times 10^{-3}}=\mathrm{f}$ $f=\frac{10^{5}}{16}=\frac{100 \times 100 \times 10}{4 \times 4}$ $=6250 \mathrm{\,Hz}$ $\frac{\Delta f}{f}=\frac{1}{2} \frac{\Delta \mu}{\mu}+\frac{\Delta \text { slope }}{\text { slope }}$ $=\frac{1}{10}+\frac{0.3}{80}=\frac{11}{80}$ $\Delta \mathrm{f}=6250 \times \frac{11}{80}=859.8 \mathrm{\,Hz}$

Question

A$6250 ± 859.8\ Hz$
B$1250 ± 32.3\ Hz$
C$6250 ± 546.9\ Hz$
D$4875 ± 287\ Hz$

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