This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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The relationship between the frequency (f) and length (L) of a vibrating string with constant tension is given by f = (1)/(2L) sqrt((T)/()), where T is tension and is linear density. This implies that f is inversely proportional to L, or f (1)/(L). To obtain a straight line graph, we should plot f against (1)/(L). First, let's calculate (1)/(L) for each data point. | Frequency f (Hz) | Length L (cm) | (1)/(L) (cm⁻¹) | f · L (Hz·cm) | |:-------------------|:----------------|:--------------------|:--------------------| | 246 | 78.1 | 0.01280 | 19212.6 | | 288 | 69.8 | 0.01433 | 20102.4 | | 320 | 62.5 | 0.01600 | 20000.0 | | 384 | 52.1 | 0.01919 | 19998.4 | | 312 | 39.2 | 0.02551 | 12230.4 | Notice that the product f · L is approximately constant for the first four points (around 20000 Hz·cm), but the last point (312 Hz, 39.2 cm) gives a significantly different product (12230.4 Hz·cm). This indicates that the last data point is an outlier. For the "best" straight line fit, we will exclude this outlier and use the first four data points for the linear regression. a) Draw a graph for the set of observation and fit the "best" straight line to them. Step 1: Prepare the data for plotting. Calculate the inverse of the length, (1)/(L), for each of the first four data points: • For L = 78.1 cm, (1)/(L) = (1)/(78.1) ≈ 0.01280 cm^-1 • For L = 69.8 cm, (1)/(L) = (1)/(69.8) ≈ 0.01433 cm^-1 • For L = 62.5 cm, (1)/(L) = (1)/(62.5) = 0.01600 cm^-1 • For L = 52.1 cm, (1)/(L) = (1)/(52.1) ≈ 0.01919 cm^-1 Step 2: Describe how to draw the graph. Plot frequency f (in Hz) on the y-axis and (1)/(L) (in cm⁻¹) on the x-axis. Mark the four data points: (0.01280, 246), (0.01433, 288), (0.01600, 320), and (0.01919, 384). Draw a straight line that best fits these four points. This line should pass close to the origin, as theoretically, f=0 when L (i.e., (1)/(L)=0). b) Find the slope and intercept of the graph. Step 1: Calculate sums for linear regression using the first four points. Let x = (1)/(L) and y = f. n = 4 x = 0.01280 + 0.01433 + 0.01600 + 0.01919 = 0.06232 y = 246 + 288 + 320 + 384 = 1238 x^2 = (0.01280)^2 + (0.01433)^2 + (0.01600)^2 + (0.01919)^2 = 0.000993445 xy = (0.01280 × 246) + (0.01433 × 288) + (0.01600 × 320) + (0.01919 × 384) = 19.766 Step 2: Calculate the slope (m) and intercept (c). The formulas for linear regression are: m = (n( xy) - ( x)( y))/(n( x^2) - ( x)^2) c = (( y)( x^2) - ( x)( xy))/(n( x^2) - ( x)^2) Substitute the calculated sums: m = (4(19.766) - (0.06232)(1238))/(4(0.000993445) - (0.06232)^2) m = (79.064 - 77.15936)/(0.00397378 - 0.0038837824) m = (1.90464)/(0.00009000) ≈ 21162.67 Hz · cm Rounding to four significant figures, the slope is 21160 Hz·cm. c = ((1238)(0.000993445) - (0.06232)(19.766))/(0.00009000) c = (1.22949011 - 1.23098012)/(0.00009000) c = (-0.00149001)/(0.00009000) ≈ -16.555 Hz Rounding to three significant figures, the intercept is -16.6 Hz. c) Find the relationship between the frequency (f) of the vibration and the length (l) of the stretched string. Step 1: State the theoretical relationship. The theoretical relationship for a vibrating string is f = k · (1)/(L), where k is a constant. Step 2: State the experimental relationship. Using the calculated slope (m) and intercept (c) from part (b), the experimental relationship is in the form f = m ((1)/(L)) + c. Substituting the values: f = 21160 ((1)/(L)) - 16.6 The relationship is f = (21160)/(L) - 16.6, where f is in Hz and L is in cm. d) What is the frequency of an unmarked fork which was in tune with a length of 45.0 cm? Step 1: Use the relationship found in part (c). Given L = 45.0 cm. f = (21160)/(L) - 16.6 Step 2: Substitute the value of L and calculate f. f = (21160)/(45.0) - 16.6 f = 470.222... - 16.6 f = 453.622... Hz Rounding to one decimal place, the frequency is 453.6 Hz. Drop the next question! 📸