A block-spring system has a maximum restoring force Fmax = 0.1 N. If theamplitude of the motion is A = 0.16 m and the mass of the block is m 100 g thenthe angular frequency w is equal to5 rad/s10 rad/s20 rad/s1.25 rad/sO 2.5 rad/sAsimple pendulum consists of a 0.8-kg bob connected to a masslessInextensible cord with a length L = 1.1 m. The bobis set into motion and itsangular displacement is given by e(t) - 0.11cos(ot), where e is in radians and t is98 m/s 2, determine the mechanical energy of this

Name: A block-spring system has a maximum restoring force Fmax = 0.1 N. If
Uploaded: 2024-03-20T06:58:04.000Z
Channel: Bartleby
Description: A block-spring system has a maximum restoring force Fmax = 0.1 N. If theamplitude of the motion is A = 0.16 m and the mass of the block is m 100 g thenthe angular frequency w is equal to5 rad/s10 rad/s20 rad/s1.25 rad/sO 2.5 rad/sAsimple pendulum co

Maximum Force, F=0.1 N Amplitude of motion, A=0.16 m Mass of block, m=100 g =0.100 kg…

Science

Physics

A block-spring system has a maximum restoring force Fmax = 0.1 N. If the amplitude of the motion is A = 0.16 m and the mass of the block is m = 100 g then %3D the angular frequency w is equal to 5 rad/s O 10 rad/s 20 rad/s 1.25 rad/s 2.5 rad/s

A block-spring system has a maximum restoring force Fmax = 0.1 N. If the amplitude of the motion is A = 0.16 m and the mass of the block is m = 100 g then %3D the angular frequency w is equal to 5 rad/s O 10 rad/s 20 rad/s 1.25 rad/s 2.5 rad/s

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter12: Oscillatory Motion

Section: Chapter Questions

Problem 7OQ: If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the...

See similar textbooks

Similar questions

If a car has a suspension system with a force constant of 5.00104 N/m , how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m?
A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.
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 .
A pendulum with a period of 2.00000 s in one location (g=9.80m/s2) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
A pendulum of length L and mass M has a spring of force constant k connected to it at a distance h below its point of suspension (Fig. P15.55). Find the frequency of vibration of the system for small values of the amplitude (small ). Assume the vertical suspension rod of length L is rigid, but ignore its mass. Figure P15.35
Show 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.
A 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.
A diver on a diving board is undergoing SHM. Her mass is 55.0 kg and the period of her motion is 0.800 s. The next diver is a male whose period of simple harmonic oscillation is 1.05 s. What is his mass if the mass of the board is negligible?
We do not need the analogy in Equation 16.30 to write expressions for the translational displacement of a pendulum bob along the circular arc s(t), translational speed v(t), and translational acceleration a(t). Show that they are given by s(t) = smax cos (smpt + ) v(t) = vmax sin (smpt + ) a(t) = amax cos(smpt + ) respectively, where smax = max with being the length of the pendulum, vmax = smax smp, and amax = smax smp2.
An automobile with a mass of 1000 kg, including passengers, settles 1.0 cm closer to the road for every additional 100 kg of passengers. It is driven with a constant horizontal component of speed 20 km/h over a washboard road with sinusoidal bumps. The amplitude and wavelength of the sine curve are 5.0 cm and 20 cm, respectively. The distance between the front and back wheels is 2.4 m. Find the amplitude of oscillation of the automobile, assuming it moves vertically as an undamped driven harmonic oscillator. Neglect the mass of the wheels and springs and assume that the wheels are always in contact with the road.
Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency , the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (c) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s.