a mass is hung from a vertical spring and allowed to come to rest or its equilibrium position. the mass is then pulled down an additional 0.25 m and released. as the mass oscillates, it completes one full cycle in 5.0 s. using the given values, correctly identify the mass's amplitude, full range of vertical motion, frequency, and period. the full range of vertical motion is the distance between the maximum and minimum heights of the mass.

a mass is hung from a vertical spring and allowed to come to rest or its equilibrium position. the mass is then pulled down an additional 0.25 m and released. as the mass oscillates, it completes one full cycle in 5.0 s. using the given values, correctly identify the mass's amplitude, full range of vertical motion, frequency, and period. the full range of vertical motion is the distance between the maximum and minimum heights of the mass.

Similar questions

A mass on the end of a spring oscillates with an amplitude of 35cm at a frequency of 20 Hz (cycles per second). At t = 0 the mass is at its equilibrium position (x = 0). a) Find the possible equations describing the position of the mass as a function of time, in the form x = A cos (ωt + φ), giving the numerical values of A, ꞷ and φ. b) What are the values of x, dx/dt, and d2x/dt2 at t = 3⁄4 sec?
In the figure below, the pendulum consists of a uniform disk with radius r = 34 cm and mass 520 g attached to a uniform rod with length L = 352 mm and mass 207 g. Calculate the period (in seconds) of oscillation. Final answer in four decimal places.
A force of 6 N will hold a spring stretched at 3 m. Suppose a 6 kg mass is attached to the spring, it is stretched to 3 m, and released with an initial velocity of 0 m/s. Assume there is no dampening. Find an equation involving y (position) and v (velocity, which is the derivative of position) that governs the motion of the mass.
The oscillatory movement of a simple pendulum is a characteristic of the regular repetition of displacements around an equilibrium position. The pendulum swings due to the restored force of gravity, which seeks to bring the mass back to the equilibrium point. The oscillatory behavior is described by a trigonometric solution, which relates the pendulum's position to time, considering its amplitude, frequency and initial phase. Statement: A simple pendulum moves according to the following question:y(t) = A.sin(ωt + φ)In this question, y(t) is the horizontal position of the pendulum, A is its amplitude, ω is its angular velocity, given by ω = 2πf, and φ is the initial phase of the movement, in radians. Since the initial phase of the movement is equal to 0 and its angular velocity is π/2 rad/s, the oscillation frequency of this pendulum is correctly given by the alternative: a) 2.0Hz b) 1.5Hz c) 1.0Hz d) 0.5Hz e) 0.25Hz
A spring (200 N/m) is attached to a mass (0.4kg) and are positioned horizontally. When the mass is at x = 2 cm it is the mass moving at 1.3 m/s. Write an equation for the position of the mass with respect to time. Assume that at t = 0 the mass is passing through equilibrium.
The period p of a pendulum, or the time it takes for the pendulum to make one complete swing, varies directly as the square root of the length L of the pendulum. If the period of a pendulum is 1.1 s when the length is 2 ft, find the period when the length is 3 ft. Round to the nearest hundredth.
An oscillator has an equation of x(t) = (0.12 m) cos( 4.2t) Assuming that w is in SI units (radians/s), find the period of oscillation in seconds with two digits of precision. Your Answer: Answer
Assume an oscillator described by the following 2nd order DE: x" + 12x' + 100x = 0 where the displacement x is function of time t. a) Find the general solution i.e. x(t). b) Calculate the period of the oscillations; what would be the period of the same oscillator if the motion was not damped? c) Given that the initial position of the oscillator is 0.5 (SI units) away from equilibrium in the +x direction and that its initial velocity is -10 (SI units); find the constants in the solution obtained in a). d) Write now your solution in Amplitude -phase form x(t) = Acos(wt – 0); 0 must be expressed in radians.
answer
A simple harmonic motion of a point mass m can be expressed byx = 0.25sin(4pt +p/6) m ( time t in ses). a)Find the amplitude for the oscillation in m. b)Find the frequency of the oscillation (in Hz). c)Find the maximum speed for the oscillation (in m/s d)Find the maximum acceleration for the oscillation (in m/s2 ) e)Find the initial position of the mass m (in m) f)Find the total energy for the oscillation (in Jules)
A spring is attached to the ceiling and pulled 7 cm down from equilibrium and released. The amplitude decreases by 12% each second. The spring oscillates 17 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t. D(t) =
A body at the end of a spring oscillates up and down forty times in 8.0 s. Calculate the period and frequency of the oscillation.