Question

A 0.45·kg mass vibrates according to the equation x = 0.45 cos (8.36t + 4.92), where...

A 0.45·kg mass vibrates according to the equation x = 0.45 cos (8.36t + 4.92), where x is in meters, and t is in seconds.

(a) Determine the amplitude.
m

(b) Determine the frequency.

Hz

(c) Determine the total energy.

J

(d) Determine the kinetic energy and potential energy when x = 0.35 m.
kinetic energy =

J
potential energy = J

Homework Answers

Answer #1

use properties of simple harmonic motion to find the required amplitude, frequency and energy as shown below

***********************************************************************************************
This concludes the answers. If there is any mistake or omission, let me know in the comments immediately and I will fix it....

Know the answer?
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
Similar Questions
A 0.55·kg mass vibrates according to the equation x = 0.41 cos (8.41t + 1.52), where...
A 0.55·kg mass vibrates according to the equation x = 0.41 cos (8.41t + 1.52), where x is in meters, and t is in seconds. (a) Determine the amplitude. m (b) Determine the frequency. Hz (c) Determine the total energy.J (d) Determine the kinetic energy and potential energy when x = 0.29 m. kinetic energy = J potential energy =  J
A 0.55·kg mass vibrates according to the equation x = 0.44 cos (8.41t + 4.83), where...
A 0.55·kg mass vibrates according to the equation x = 0.44 cos (8.41t + 4.83), where x is in meters, and t is in seconds. (a) Determine the amplitude.   m (b) Determine the frequency. Hz (c) Determine the total energy. J (d) Determine the kinetic energy and potential energy when x = 0.32 m. kinetic energy = J potential energy =   J
A 0.50 kg mass vibrates according to the equation x=0.50cos6.80t, where x is in meters and...
A 0.50 kg mass vibrates according to the equation x=0.50cos6.80t, where x is in meters and t is in seconds. A) Determine the amplitude. B) Determine the frequency. C) Determine the total energy D) Determine the kinetic energy when x = 0.36 m . E) Determine the potential energy when x = 0.36 m .
A 1.00 kgkg mass oscillates according to the equation x=0.700cos8.60tx=0.700cos⁡8.60t, where xx is in meters and...
A 1.00 kgkg mass oscillates according to the equation x=0.700cos8.60tx=0.700cos⁡8.60t, where xx is in meters and tt is in seconds Part D Determine the kinetic energy when x = 0.490 m Determine the potential energy when x = 0.490 mm .
A 0.69 kg mass at the end of a spring vibrates 4.0 times per second with...
A 0.69 kg mass at the end of a spring vibrates 4.0 times per second with an amplitude of 0.13 m . A) Determine the velocity when it passes the equilibrium point. B) Determine the velocity when it is .11 m from equilibrium. C) Determine the total energy of the system. D) Determine the equation describing the motion of the mass, assuming that x was a maximum at t=0.
A mass on a spring with constant 3.45 N/m vibrates with position given by the equation...
A mass on a spring with constant 3.45 N/m vibrates with position given by the equation x = (5.00 cm) cos(4.40t rad/s). (a) During the first cycle, for 0 < t < 1.43 s, just when is the system's potential energy of the system changing most rapidly into kinetic energy? first time s second time s (b) What is the maximum rate of energy transformation? mW
A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrates in...
A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.2 cm. the maximum value of its speed is 54.6 WHAT IS THE MAXIMUM VALUE OF IT'S ACCELERATION? QUESTION 2 A 45.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 7.00 cm on a frictionless, horizontal surface. the total energy of the system is 98 the speed of...
The position of a certain system with mass of 10 kg exhibits simple harmonic motion, where...
The position of a certain system with mass of 10 kg exhibits simple harmonic motion, where x(t) = 20 cos(15.2t + ?/4) and is in units of meters. (a) What is the total energy of the system (let the potential energy be zero at the equilibrium position)? (b) At t = 0, what is the potential energy? (c) Kinetic energy? Please answer with the concept of SHM
A particle with mass 1.23 kg oscillates horizontally at the end of a horizontal spring. A...
A particle with mass 1.23 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.847 m and a duration of 125 s for 69 cycles of oscillation. Find the frequency, ?, the speed at the equilibrium position, ?max, the spring constant, ?, the potential energy at an endpoint, ?max, the potential energy when the particle is located 41.3% of the amplitude away from the equiliibrium position, ?, and the kinetic energy, ?, and...
13.7 The equation for the position as a function of time for an oscillating spring is...
13.7 The equation for the position as a function of time for an oscillating spring is given by x  30cmcos 25t where x is in centimeters when t is in seconds. a) What is the frequency? b) If the mass on the spring is 1.2 kg, what is the spring constant of the spring? c) What is the position at t = 0.025 s? d) What is the position at t = 0.09 s ? 13.8 The maximum potential...