iPad 8:35 PM X standing wave docx Data presentation Mass of string 30.6656g Tota

Question

iPad 8:35 PM X standing wave docx Data presentation Mass of string 30.6656g Total length of sting E208cm Linear mass density 6.2 10 kg/m Length of string between vibrator and pulley L 120 khz Data table Number of Suspended Total Force Measured Wavelength M length Ln A measured (Kg for N loops 1 0943 F1 925 Lil 1.15 25 6.25 304 2 0.500 F2 4.90 L2 1.07 2.14 4.5796 2.2136 0.450 F4 4.4145 LA 0.70 1.94 3.7636 2.1011 0.400 F5 3924 L5 0915 183 3.3469 1.9810 6 0.350 F6 34335 L6 0.840 1.63 26569 1.8530 0.250 24555 0.790 1 48 2.1904 1.5664 0.200 F8 4.962 L8 0.605 1.28 1.6384 1 4001 Slope of graph 126. 66khz Computed frequency f 166.95 Accepted frequency 120khz Percentage error *39.125% f slope eqn 1. slo 126.66 1 6.2 10 Slope 9.9465 Using equation 1 slope f xu If the frequency is 60 hz,

Explanation / Answer

Note:- Since you have already done the first question and it is correct also, so I am doing only 2nd Question.

Solution:- a) We know that f = SQRT (slope/u)

So squaring both sides

f^2 = slope/u

slope = u* f^2

So if frequnecy is halfed that is from 120 to 60 then slope will decrease much. It will decrease in square of f.

b) If the scattered data points far from the straight line on the graph then data for these points can help in calculating the frequency by the following problem:

f = SQRT (slope / u) = SQRT (0.1827/6.2*10^-4) =17.167

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