How does the mass on a horizontal spring behave?

[mattmannmf]

A 53 gram mass is attached to a massless spring and allowed to oscillate around an equilibrium according to:
y(t) = 1.2*sin( 3.1415*t ) where y is measured in meters and t in seconds

(a) What is the spring constant in N/m ?

(b) What is the total Mechanical Energy in the mass/spring system?

(c) What is the maximum Kinetic Energy of the mass?

(d) What is the maximum velocity of the mass in m/s ?

 

[LowlyPion]

What are your thoughts on how to approach it?

 

[nlightner]

(a) The spring constant can be calculated using the equation k = mω^2, where m is the mass in kg and ω is the angular frequency in radians per second. In this case, the mass is 53 grams, or 0.053 kg, and the angular frequency is 3.1415 radians per second. Plugging these values into the equation, we get k = (0.053 kg)(3.1415 rad/s)^2 = 0.53 N/m.

(b) The total mechanical energy in the system can be calculated using the equation E = (1/2)kA^2, where k is the spring constant and A is the amplitude of oscillation. In this case, the amplitude can be found by taking the absolute value of the coefficient in front of the sine function, which is 1.2 meters. Plugging in the values, we get E = (1/2)(0.53 N/m)(1.2 m)^2 = 0.38 J.

(c) The maximum kinetic energy of the mass occurs at the equilibrium point, where all of the potential energy is converted into kinetic energy. The maximum kinetic energy can be calculated using the equation KE = (1/2)mv^2, where m is the mass in kg and v is the velocity in meters per second. In this case, the mass is 0.053 kg and the velocity can be found by taking the derivative of the position equation with respect to time. This gives us v = 1.2*3.1415*cos(3.1415*t) = 3.7695*cos(3.1415*t) m/s. At the equilibrium point, t = 0, so the maximum velocity is 3.7695 m/s. Plugging this into the equation, we get KE = (1/2)(0.053 kg)(3.7695 m/s)^2 = 0.38 J.

(d) The maximum velocity can be found using the same equation as in part (c). At the maximum displacement, the velocity is 0, so we can set the position equation equal to 0 and solve for t. This gives us t = 0.3183 seconds. Plugging this value into the velocity equation, we get v = 3.7695*cos(3.1415*0.3183) = 3.7695 m