Question

Pre-Lab Assignment for Experiment 11: The frequency and the half wavelength d of a standing wave on a string are related by = 2 V / u. using the tension Fand the linear mass density. In this experiment, you will vary the tension in the string, and measured for several different standing waves, with the goal of finding the frequency f. While one could find a value for ffrom each single data point, it is often a better strategy to use the ensemble of all your data and transform them into a linear plot. Thus, the random errors made in each single point become diminished, and the experimental result is representative of all of your data points The equation above can be written as F=(4ufd . In a plot of F versus d. data points should lie along a straight line of slope (41 ) as in the example below. Find slope and uncertainty of the slope for the best-fit line in the graph below. Use techniques discussed in earlier experiments and Appendix D. Find the respective frequency f, if p=0.315 g/m. dim) Ad(m) AF (N) of (m) 0.325 4d (m) 0.017 0.570 0.015 0.34 0.423 0.015 0.179 0.013 F(N) 5.47 3.04 1.96 1.34 0.338 0.015 0.114 0.010 0.10 0.10 0.03 0.283 0.015 0.080 0.009

Answer 1

We have the expression relatinfg frequency and half wave length of the string related by,

$f=\frac{1}{2d}\sqrt{\frac{F}{\mu }}$

Where =tension in the string in Newtons(N)

And,$\mu$=Linear mass density(kg/m)

Also from the equation,on squaring both sides,

$f^{2}=\frac{F}{4d^{2}\mu }$

Or ,ie

F = (4uf?)*

So,From the table of the observation of the experiment,given in question,we find the different values of $d^{2}$ by varying the tension in the string .

On plotting the graph of  $d^{2}(m^{2})$ vs :from the given values in the table,with $d^{2}(m^{2})$ on x axis and on y axis.

C D E F G H I J B 1 d2(m2) FIN) 2 0.325 5.47 3 0.179 3.04 0.114 1.96 0.08 1.34 d2(m2) v/s F(N) y = 16.781x +0.0243 5.47 F(N) 304 1.96 1.34 0.05 0.1 0.25 0.3 0.35 0.15 0.2 d2 (m2)

Here the equation of the line of fit found to be,as shown,$y=16.781x+0.0243$

It is the equation of the line,$y=mx+c$

So,the slope of the line fit,$m=16.781N/m^{2}$

So,here we have the slope of the line from the equation that,

m=2= 4uf? = 16.781N/m?

So,we have $4\mu f^{2}= 16.781N/m^{2}$

Also we have the slope,$m=\frac{F}{d^{2}}$

For finding the uncertainity in slope,relative uncertainity in slope,$\frac{\Delta m}{m}=\frac{\Delta F}{F}+\frac{\Delta (d^{2})}{d^{2}}$

Here,the mean value of $d^{2}=\frac{0.325+0.179+0.114+0.08}{4}=0.1745m^{2}$

And the mean value of $\Delta d^{2}=\frac{0.017+0.013+0.01+0.009}{4}=0.01225m^{2}$

mean value of $\Delta F=\frac{0.34+0.10+0.10+0.03}{4}=0.1425N$

Mean value of $F=\frac{5.47+3.04+1.96+1.34}{4}=2.9525N$

So,ie,$\frac{\Delta m}{m}=\frac{0.1425}{2.9525}+\frac{0.01225}{0.1745}=0.1185$

Also we have from the graph,slope,$m=16.781N/m^{2}$

So,eqn becomes,$\Delta m=0.1185*m=0.1185*16.781=1.9885N/m^{2}$

uncertainity of slope,$\Delta m=1.9885N/m^{2}$

So,the slope of the graph with uncertainity,$m\pm \Delta m=16.781\pm 1.9885N/m^{2}$

We have given for the value of linear mass density,$\mu =0.315g/m^{2}=3.15*10^{-4}kg/m^{2}$

So,for this value of $\mu$ ,the value of frequency can be find from slope,$m=4\mu f^{2}$

So,$4\mu f^{2}=m=16.781N/m^{2}$

Then,frequency,$f=\sqrt{\frac{m}{4\mu }}=\sqrt{\frac{16.781}{4*3.15*10^{-4}}}=115.404Hz$

So,frequency,$f=115.404Hz$

Here to find uncertainity in frequency,$\frac{\Delta m}{m}=2*\frac{\Delta f}{f}$

So,$\Delta f=\frac{\Delta m*f}{2m}=\frac{1.9885*115.404}{2*16.781}=6.837Hz$

And the frequency,$f\pm \Delta f=115.40\pm 6.837Hz$

Or the value of frequency,$f=115.404Hz$

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