Concept explainers
(a)
The force constant of the spring.
(a)
Answer to Problem 18P
The force constant of the spring is
Explanation of Solution
Write the expression to calculate the force constant of the spring.
Here,
Conclusion:
Substitute
Thus, the force constant of the spring is
(b)
The frequency of oscillations.
(b)
Answer to Problem 18P
The frequency of oscillations is
Explanation of Solution
Write the formula to calculate the angular frequency for the object.
Here,
Write the formula to calculate the frequency of oscillations.
Here,
Conclusion:
Substitute
Substitute
Thus, the frequency of oscillations is
(c)
The value for the maximum speed of the object.
(c)
Answer to Problem 18P
The value for the maximum speed of the object is
Explanation of Solution
Write the expression to calculate the maximum speed of the object.
Here,
Conclusion:
Substitute
Thus, the value for the maximum speed of the object is
(d)
The position at which the maximum speed occur.
(d)
Answer to Problem 18P
The position at which the maximum speed occur is
Explanation of Solution
The maximum speed of a
Thus, the position at which the maximum speed occur is
(e)
The maximum acceleration of the object.
(e)
Answer to Problem 18P
The maximum acceleration of the object is
Explanation of Solution
Write the expression to calculate the maximum acceleration of the object.
Here,
Conclusion:
Substitute
Thus, the maximum acceleration of the object is
(f)
The position at which the maximum acceleration occur.
(f)
Answer to Problem 18P
The position at which the maximum acceleration occur is
Explanation of Solution
The object attains its maximum acceleration when the direction of the object is reversed. The object reverses its direction when it is at maximum distance from the equilibrium x position.
The object reverses its direction at the maximum position of
Thus, the position at which the maximum acceleration occur is
(g)
The total energy of the oscillating system.
(g)
Answer to Problem 18P
The total energy of the oscillating system is
Explanation of Solution
Write the expression to calculate the total energy of the oscillating system.
Here,
Conclusion:
Substitute
Thus, the total energy of the oscillating system is
(h)
The speed of the object when its position is one third of maximum value.
(h)
Answer to Problem 18P
The speed of the object when its position is one third of maximum value is
Explanation of Solution
Write the expression to calculate the speed of the object at
Conclusion:
Substitute
Thus, the speed of the object when its position is one third of maximum value is
(i)
The acceleration of the object when its position is one third of maximum value.
(i)
Answer to Problem 18P
The acceleration of the object when its position is one third of maximum value is
Explanation of Solution
Write the expression to calculate the speed of the object at
Conclusion:
Substitute
Thus, the acceleration of the object when its position is one third of maximum value is
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Chapter 12 Solutions
Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term
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- Four people, each with a mass of 72.4 kg, are in a car with a mass of 1 130 kg. An earthquake strikes. The vertical oscillations of the ground surface make the car bounce up and down on its suspension springs, but the driver manages to pull off the road and stop. When the frequency of the shaking is 1.80 Hz, the car exhibits a maximum amplitude of vibration. The earthquake ends and the four people leave the car as fast as they can. By what distance does the cars undamaged suspension lift the cars body as the people get out?arrow_forwardA spherical bob of mass m and radius R is suspended from a fixed point by a rigid rod of negligible mass whose length from the point of support to the center of the bob is L (Fig. P16.75). Find the period of small oscillation. N The frequency of a physical pendulum comprising a nonuniform rod of mass 1.25 kg pivoted at one end is observed to be 0.667 Hz. The center of mass of the rod is 40.0 cm below the pivot point. What is the rotational inertia of the pendulum around its pivot point?arrow_forwardWhen a block of mass M, connected to the end of a spring of mass ms = 7.40 g and force constant k, is set into simple harmonic motion, the period of its motion is T=2M+(ms/3)k A two-part experiment is conducted with the use of blocks of various masses suspended vertically from the spring as shown in Figure P15.76. (a) Static extensions of 17.0, 29.3, 35.3, 41.3, 47.1, and 49.3 cm are measured for M values of 20.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, respectively. Construct a graph of Mg versus x and perform a linear least-squares fit to the data. (b) From the slope of your graph, determine a value for k for this spring. (c) The system is now set into simple harmonic motion, and periods are measured with a stopwatch. With M = 80.0 g, the total time interval required for ten oscillations is measured to be 13.41 s. The experiment is repeated with M values of 70.0, 60.0, 50.0, 40.0, and 20.0 g, with corresponding time intervals for ten oscillations of 12.52, 11.67, 10.67, 9.62, and 7.03 s. Make a table of these masses and times. (d) Compute the experimental value for T from each of these measurements. (e) Plot a graph of T2 versus M and (f) determine a value for k from the slope of the linear least-squares fit through the data points. (g) Compare this value of k with that obtained in part (b). (h) Obtain a value for ms from your graph and compare it with the given value of 7.40 g.arrow_forward
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