A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resulting oscillation has a period of 0.750 s and an amplitude of 15.0 cm. Determine the oscillation frequency, the spring constant, and the speed of the mass when it is halfway to the equilibrium position.

the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

Explanation:

Given that;

mass of the object m = 1.20 kg

period of oscillation = 0.750 s

Amplitude ( A/x) = 15.0 cm = 0.15 m

now;

a) Determine the oscillation frequency;

oscillation frequency f = 1/T

we substitute

f = 1 / 0.750 s

f = 1.33 Hz

Therefore, the oscillation frequency is 1.33 Hz

b) Determine the spring constant;

we solve for spring constant from the following expression;

T = 2π√(m/k)

k = 4π²m / T²

so we substitute

k = (4π² × 1.20) / (0.750)²

k = 47.3741 / 0.5625

k =  84.22 N/m

Therefore, the spring constant is 84.22 N/m

c) determine the speed of the mass when it is halfway to the equilibrium position

So, at equilibrium, the energy is equal to K.E

such that;

1/2mv² = 1/2kx²

mv² = kx²

v² = kx² / m

v = √( kx²/m)

we substitute

v = √( 84.22×(0.15 m)²/ 1.2 )

v = √( 1.89495 / 1.2 )

v = √ 1.579125

v = 1.26 m/s  

Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

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  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.