Q90P - University Physics with Modern Physics

Question

You hang various masses m from the end of a vertical, 0.250-kg spring that obeys Hooke’s law and is tapered, which means the diameter changes along the length of the spring. Since the mass of the spring is not negligible, you must replace m in the equation \(T = 2\pi \sqrt {\frac{m}{k}} \) \(m + {m_{eff}}\) where \({m_{eff}}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass m and measure the time for 10 complete oscillations, obtaining these data:

m(kg)	0.100	0.200	0.300	0.400	0.500
Time (s)	8.7	10.5	12.2	13.9	15.1

 

(a) Graph the square of the period T versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring’s effective mass. (d) What fraction is \({m_{eff}}\) of the spring’s mass? (e) If a 0.450-kg mass oscillates on the end of the spring, find its period, frequency, and angular frequency

Step-by-Step Solution

Answer

(a)(i) The graph is drawn below. (ii) \({T^2} = \left( {3.878\;{{\rm{s}}^{\rm{2}}}{\rm{/kg}}} \right)m + 0.3492\;{{\rm{s}}^{\rm{2}}}\)

Step 1: Given Data

m(kg)	0.100	0.200	0.300	0.400	0.500
Time (s)	8.7	10.5	12.2	13.9	15.1

Step 2: Concept

The time period is determined by finding the one complete duration of oscillation.

Step 3(a) (i): Graph the square of the period T versus the mass suspended from the spring

m(kg)	0.100	0.200	0.300	0.400	0.500
Time (s)	8.7	10.5	12.2	13.9	15.1
Time Period (s) = (Time/10)(s)	0.87	1.05	1.22	1.39	1.51

 

Mass (kg)	0.100	0.200	0.300	0.400	0.500
\({T^2}({s^2})\)	0.7569	1.1025	1.4884	1.9321	2.2801

 

Plot the graph between mass and \({T^2}\) 

Hence the graph is drawn.

Step 3(a) (ii): Find the straight line of best fit

The best fit line is given by \({T^2} = \left( {3.878\;{{\rm{s}}^{\rm{2}}}{\rm{/kg}}} \right)m + 0.3492\;{{\rm{s}}^{\rm{2}}}\) 

Hence the straight line of best fit is \({T^2} = \left( {3.878\;{{\rm{s}}^{\rm{2}}}{\rm{/kg}}} \right)m + 0.3492\;{{\rm{s}}^{\rm{2}}}\)