Question

QUESTION 33 In simple harmonic motion, the speed is zero at that point in the cycle when the acceleration is zero. the potent

3 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
QUESTION 33 In simple harmonic motion, the speed is zero at that point in the cycle...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Energy in simple harmonic motion

    A 0.500-kg glider, attached to the end of an ideal spring with force constant k=450N/m, undergoes simple harmonic motion with an amplitude 0.040m. Compute a) themaximum speed of the glider; b) the speed of the glider when it is at x= -0.015m; c)the magnitude of the maximum acceleration of the glider; d)the acceleration of theglider at x= -0.015m; e) the total mechanical energy of the glider at any point in its motion.

  • In simple harmonic motion, the speed is greatest at that point in the cycle whenA) the...

    In simple harmonic motion, the speed is greatest at that point in the cycle whenA) the magnitude of the acceleration is a maximum.B) the displacement is a maximum.C) the magnitude of the acceleration is a minimum.D) the potential energy is a maximum.E) the kinetic energy is a minimum.

  • Simple Harmonic Motion. Effective Spring Constant: In Part I you measured the keffective of the two...

    Simple Harmonic Motion. Effective Spring Constant: In Part I you measured the keffective of the two springs acting together. If the two springs had k1 and k2 individually, how would they combine to get keff? Systemic Error: Leveling Air Track: We level the air track in this lab because it's good lab procedure in general, but in fact in this lab a not-level (but still straight) air track shouldn't change any of our results. Explain why this is true (for...

  • A 0.500-kg glider on an air track is attached to the end of an ideal spring...

    A 0.500-kg glider on an air track is attached to the end of an ideal spring with force constant 470 N/m; it undergoes simple harmonic motion with an amplitude of 5.00×10^−2 m. A. Calculate the maximum speed of the glider. B. Calculate the speed of the glider when it is at −1.40×10^−2 m. C. Calculate the magnitude of the maximum acceleration of the glider. D. Calculate the acceleration of the glider at −1.40×10^−2 m. E. Calculate the total mechanical energy...

  • Energy in simple harmonic motion A 2.90 kg object oscillates with simple harmonic motion on a...

    Energy in simple harmonic motion A 2.90 kg object oscillates with simple harmonic motion on a spring of force constant 600 N/m. The maximum speed is 0.800 m/s. A) What is the total energy of the object and the spring? B) What is the maximum amplitude of the oscillation?

  • Review Constants Let's begin with a straightforward example of simple harmonic motion (SHM). A spring is...

    Review Constants Let's begin with a straightforward example of simple harmonic motion (SHM). A spring is mounted horizontally on an air track as in (Figure 1), with the left end held stationary. We attach a spring balance to the free end of the spring, pull toward the right, and measure the elongation. We determine that the stretching force is proportional to the displacement and that a force of 60 N causes an elongation of 0.030 m. We remove the spring...

  • Question 7 1 pts A block attached to a spring is undergoing simple harmonic motion. At...

    Question 7 1 pts A block attached to a spring is undergoing simple harmonic motion. At one point in its motion, its kinetic energy is 5 J and its potential energy is 3 J. When the block reaches the point of maximum displacement from equilibrium, the kinetic and potential energies are: K-0 and U--8 Previous Submit Quiz No new data to save. Last checked at 10:39am

  • A mass of 397 g is attached to a spring and set into simple harmonic motion...

    A mass of 397 g is attached to a spring and set into simple harmonic motion with a period of 0.246 s. If the total energy of the oscillating system is 5.94 J, determine the following. (a) maximum speed of the object 6.49 When is the total energy of the mass-spring system equal to the kinetic energy of the mass? m/s (b) force constant N/m (c) amplitude of the motion Additional Materials Reading

  • 1. A simple harmonic motion of an object of mass m = 11 kg attached with...

    1. A simple harmonic motion of an object of mass m = 11 kg attached with a spring is represented as time vs displacement graph in the following figure. Find the following parameters. (a) Amplitude = (b) Time Period = ( time for 1 wavelength distance) (c) Frequency = (d) Spring Constant = (e) Angular frequency = (f) Maximum Potential Energy stored in the spring (g) Maximum Kinetic Energy of the block (h) total energy of the spring -block system

  • 1. A simple harmonic motion of an object of mass m = 11 kg attached with...

    1. A simple harmonic motion of an object of mass m = 11 kg attached with a spring is represented as time vs displacement graph in the following figure. Find the following parameters. ТАЛААР (a) Amplitude = (b) Time Period =( time for 1 wavelength distance) (c) Frequency = (d) Spring Constant = (e) Angular frequency = (1) Maximum Potential Energy stored in the spring (g) Maximum Kinetic Energy of the block (h) total energy of the spring -block system

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT