Question

An object weighing 44.1 N hangs from a vertical massless ideal spring. When set in vertical motion, the object obeys the equation y(t) = (6.20 cm) cos[(2.74 rad/s)t ‐ 1.40]. . (a) Find the time for this object to vibrate one complete cycle. . (b) What are the maximum speed and maximum acceleration of the object. . (c) What is the TOTAL distance the object moves through in one cycle. . (d) Find the maximum kinetic energy of the object. . (e) What is the spring constant of the spring.

Answer 1

To find the solutions to the given SHM question, we used the principles of simple harmonic motion to calculate the period, maximum speed and acceleration, total distance moved in one cycle, maximum kinetic energy, and spring constant, using the provided equation of motion for the object attached to the massless springs.

Step-by-step explanation:

The equation provided describes simple harmonic motion (SHM) of the object attached to the spring. To solve the problem, we shall use the information given in the equation y(t) = (6.20 cm) cos[(2.74 rad/s)t − 1.40] and apply the principles of SHM.

The time for one complete cycle, also known as the period (T), is the reciprocal of the frequency (f). Since we're given the angular frequency ω (2.74 rad/s), we can first find the frequency using the formula ω = 2πf, and then find the period using T = 1/f.

The maximum speed (v_max) occurs when the object passes through the equilibrium position, and its value can be calculated using v_max = ωA, where A is the amplitude. The maximum acceleration (a_max) occurs at the maximum displacement and can be calculated using a_max = ω²A.

The total distance moved in one cycle is four times the amplitude since the object moves from one extreme to the equilibrium, to the other extreme, and back to the equilibrium position. The maximum kinetic energy (KE_max) of the object is given when all potential energy is converted into kinetic energy at the equilibrium position, and it can be calculated using KE_max = 0.5mv_max².

The spring constant (k) can be determined using the weight of the object (W = mg = 44.1 N) and the amplitude (A), with the relationship k = W/A.

Answer 2

The time for one vibration cycle is found using the period formula for SHM, whereas the maximum speed and acceleration are derived using the amplitude and angular frequency. The total distance is four times the amplitude, and the maximum kinetic energy is calculated with the object's mass and maximum speed. Lastly, the spring constant is deduced from the angular frequency and the mass.

Step-by-step explanation:

To solve this physics problem involving an object in simple harmonic motion (SHM) attached to a spring, we'll go through each part of the question step by step.

Time for One Complete Cycle

(a) The time for one complete vibration cycle, or the period (T), can be calculated for SHM using the formula T = 2π/ω, where ω is the angular frequency given as 2.74 rad/s. Plugging in the value, we find T = 2π/2.74 s.

Maximum Speed and Acceleration

(b) Maximum speed (v_max) and maximum acceleration (a_max) for an object in SHM can be calculated using the formulas v_max = ωA and a_max = ω^2A, where A is the amplitude of the motion (6.20 cm).

Total Distance in One Cycle

(c) The total distance an object moves in one cycle of SHM is given by 4 times the amplitude.

Maximum Kinetic Energy

(d) The maximum kinetic energy (KE_max) can be calculated using the formula KE_max = 0.5mv_max^2, where m is the mass calculated from the weight (44.1 N) using the gravitational acceleration (g = 9.81 m/s^2).

Spring Constant

(e) The spring constant (k) can be found using the formula ω^2 = k/m, rearranging this gives k = mω^2.