Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305714892
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 15, Problem 15.44P

A small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three lengths. For lengths of 1.000 m, 0.750 m, and 0.500 m, total time intervals for 50 oscillations of 99.8 s, 86.6 s, and 71.1 s are measured with a stopwatch, (a) Determine the period of motion for each length, (b) Determine the mean value of g obtained from these three independent measurements and compare it with the accepted value, (c) Plot T2 versus L and obtain a value tor g from the slope of your best-fit straight-line graph. (d) Compare the value found in pan (c) with that obtained in pan (b).

(a)

Expert Solution
Check Mark
To determine

The period of motion for each length of simple pendulum.

Answer to Problem 15.44P

The period of motion for the length 1.00m of simple pendulum is 1.996s , the period of motion for the length 0.75m of simple pendulum is 1.732s and the period of motion for the length 0.50m of simple pendulum is 1.422s .

Explanation of Solution

Given info: The lengths of the simple pendulum are 1.00m , 0.75m , 0.50m and the total time intervals for 50oscillations of 99.8s , 86.6s , 71.1s are measured with a stopwatch.

For the 1.00m length of the pendulum, the total time interval of 50oscillations is 99.8s , so the period of per oscillation is the ration of total time interval and the number of oscillations,

T1=99.8s50=1.996s

Here,

T1 is the period of per oscillation of 1.00m length of the pendulum.

Thus, the period for the 1.00m length of the pendulum is 1.996s .

For the 0.75m length of the pendulum, the total time interval of 50oscillations is 86.6s , so the period of per oscillation is the ration of total time interval and the number of oscillations,

T2=86.6s50=1.732s

Here,

T2 is the period of per oscillation of 0.75m length of the pendulum.

Thus, the period for the 0.75m length of the pendulum is 1.732s .

For the 0.50m length of the pendulum, the total time interval of 50oscillations is 71.1s , so the period of per oscillation is the ration of total time interval and the number of oscillations,

T3=71.1s50=1.422s

Here,

T3 is the period of per oscillation of 0.50m length of the pendulum.

Thus, the period for the 0.50m length of the pendulum is 1.422s .

Conclusion:

Therefore, the period of motion for the length 1.00m of simple pendulum is 1.996s , the period of motion for the length 0.75m of simple pendulum is 1.732s and the period of motion for the length 0.50m of simple pendulum is 1.422s .

(b)

Expert Solution
Check Mark
To determine

The mean value of g obtained from these three independent measurements and compare it with the accepted values.

Answer to Problem 15.44P

The mean value of g for the length 1.00m of simple pendulum is 9.9m/s2 , the mean value of g for the length 0.75m of simple pendulum is 9.87m/s2 and the mean value of g for the length 0.50m of simple pendulum is 9.76m/s2 .

Explanation of Solution

Given info: The lengths of the simple pendulum are 1.00m , 0.75m , 0.50m and the total time intervals for 50oscillations of 99.8s , 86.6s , 71.1s are measured with a stopwatch.

The period of the oscillation of the pendulum is,

T=2πLg

Here,

T is the period of the oscillations.

L is the length of the pendulum.

g is the acceleration due to gravity.

Take square on the both sides and calculate the g from the above equation,

T2=(2πLg)2T2=4π2(Lg)g=4π2(LT2) (1)

Substitute 1.00m for L and 1.996s for T in above equation.

g=4π2(1.00m(1.996s)2)=4π2(0.25m/s2)=9.9m/s2

Thus, the mean value of g for the length 1.00m of simple pendulum is 9.9m/s2 .

Substitute 0.75m for L and 1.732s for T in above equation.

g=4π2(0.75m(1.732s)2)=4π2(0.433m/s2)=9.87m/s2

Thus, the mean value of g for the length 0.75m of simple pendulum is 9.87m/s2 .

Substitute 0.50m for L and 1.422s for T in above equation.

g=4π2(0.50m(1.422s)2)=4π2(0.351m/s2)=9.76m/s2

Thus, the mean value of g for the length 0.50m of simple pendulum is 9.76m/s2 .

Conclusion:

Therefore, the mean value of g for the length 1.00m of simple pendulum is 9.9m/s2 , the mean value of g for the length 0.75m of simple pendulum is 9.87m/s2 and the mean value of g for the length 0.50m of simple pendulum is 9.76m/s2 .

(c)

Expert Solution
Check Mark
To determine

To draw: The graph of T2 versus L and find the value of g from the slope of best-fit straight line graph.

Answer to Problem 15.44P

The graph of T2 versus L is shown in below figure,

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term, Chapter 15, Problem 15.44P , additional homework tip  1

Figure (1)

The value of the g from the slope of the above graph is 10.1m/s2 .

Explanation of Solution

Given info: The lengths of the simple pendulum are 1.00m , 0.75m , 0.50m and the total time intervals for 50oscillations of 99.8s , 86.6s , 71.1s are measured with a stopwatch.

In the part (a), the periods of different given length of pendulum are calculated. Make a table of square of periods T2 for given three lengths of pendulum.

L T2
1.00m (1.996s)2=3.96s2
0.75m (1.732s)2=2.99s2
0.50m (1.422s)2=2.022s2

The above table gives ordered pairs.

Take the ordered pairs from above given table and join them by a straight line and plot the graph of T2 versus L ,

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term, Chapter 15, Problem 15.44P , additional homework tip  2

Figure (1)

The Figure (1) shows the graph of T2 versus L as straight line.

From the above graph, the slope of the line is,

m=y2y1x2x1

Here,

m is the slope of the line.

x1,x2 are the two points at x -axis.

y1,y2 are the two points at y -axis.

Substitute 1.00m for x1 , 0.75m for x2 , 3.96s2 for y1 and 2.99s2 for y2 in above equation.

m=2.99s23.96s21.00m0.75m=0.97s20.25m=3.88s2/m

From the equation of the period of the pendulum,

T2=4π2Lg

The slope of T2 versus L is 4π2g .

So, 4π2g is equal to the 3.88s2/m slope of the graph.

4π2g=3.88s2/mg=4π23.88s2/m=10.1m/s2

(d)

Expert Solution
Check Mark
To determine

The comparison of values of g from part (c) and from part (b).

Answer to Problem 15.44P

The value of g from part (c) is 10.1m/s2 and the average value of g from part (b) of different length of the pendulum is 9.85m/s2 and there is no difference between them.

Explanation of Solution

Given info: The lengths of the simple pendulum are 1.00m , 0.75m , 0.50m and the total time intervals for 50oscillations of 99.8s , 86.6s , 71.1s are measured with a stopwatch.

From part (c) the value of g is 10.1m/s2 and the average value of g from part (b) of different lengths of the pendulum is 9.85m/s2 .

Both values of g have not major difference, both are approximately equal.

Conclusion:

Therefore, the value of g from part (c) is 10.1m/s2 and the average value of g from part (b) of different length of the pendulum is 9.85m/s2 and there is no difference between them.

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Chapter 15 Solutions

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term

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