Concept explainers
Change in frequency of simple pendulum on increasing its length by two times.
Answer to Problem 1OQ
Option (d). It becomes
Explanation of Solution
Write the equation to find the time period of simple pendulum.
Here,
Write the relation between frequency and
Here,
Rewrite the above expression by substituting
Rewrite the above equation to find the new frequency.
Here,
Rewrite the above equation by substituting
Conclusion:
It is found that as length of pendulum is increased by two times, frequency of oscillation gets decreases by
Therefore, options (a), (b), (c), and (e) are incorrect.
Want to see more full solutions like this?
Chapter 15 Solutions
Physics for Scientists and Engineers with Modern Physics, Technology Update
- Show that angular frequency of a physical pendulum phy=mgrCM/I (Eq. 16.33) equals the angular frequency of a simple pendulum smp=g/, (Eq. 16.29) in the case of a particle at the end of a string of length .arrow_forwardA 500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm.arrow_forwardA grandfather clock has a pendulum length of 0.7 m and mass bob of 0.4 kg. A mass of 2 kg falls 0.8 m in seven days to keep the amplitude (from equilibrium) of the pendulum oscillation steady at 0.03 rad. What is the Q of the system?arrow_forward
- Which of the following statements is not true regarding a massspring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constant. (b) The energy of the system is continually transformed between kinetic and potential energy. (c) The total energy of the system is proportional to the square of the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maximum value when the mass passes through the equilibrium position.arrow_forwardIn an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x=5.00cos(2t+6) where x is in centimeters and t is in seconds. At t = 0, find (a) the position of the piston, (b) its velocity, and (c) its acceleration. Find (d) the period and (e) the amplitude of the motion.arrow_forwardIf the amplitude of a damped oscillator decreases to 1/e of its initial value after n periods, show that the frequency of the oscillator must be approximately [1 − (8π2n2)−1] times the frequency of the corresponding undamped oscillator.arrow_forward
- A block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes simple harmonic motion with an amplitude of 10.0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 30.0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block.arrow_forwardA blockspring system oscillates with an amplitude of 3.50 cm. The spring constant is 250 N/m and the mass of the block is 0.500 kg. Determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration.arrow_forwardA 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system and (b) the speed of the object when its position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when its position is 3.00 cm.arrow_forward
- The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?arrow_forwardShow that the time rate of change of mechanical energy for a damped, undriven oscillator is given by dE/dt = bv2 and hence is always negative. To do so, differentiate the expression for the mechanical energy of an oscillator, E=12mv2+12kx2, and use Equation 15.51.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
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- Classical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning