Aspring has a 12 cm length. when a 200-g mass is hung from the spring, it extends to 27 cm. the hanging mass were pulled downward a further 5 cm. what is the time-dependent function of the position, in centimeters, of the mass, assuming that the phase angle latex: \phi=0ϕ=0?

latex: x(t)=15\cos{(8.08\textrm{ }t)}x(t)=15cos⁡(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.

latex: x(t)=5\cos{(8.08\textrm{ }t)}x(t)=5cos⁡(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.

latex: x(t)=5\cos{(1.29\textrm{ }t)}x(t)=5cos⁡(1.29 t)x(t)=5cos⁡(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.

latex: x(t)=15\cos{(1.29\textrm{ }t)}x(t)=15cos⁡(1.29 t)x(t)=15cos⁡(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.

The equation of the time-dependent function of the position is

(b) is correct option.

Explanation:

Given that,

Length = 12 cm

Mass = 200 g

Extend distance = 27 cm

Distance = 5 cm

Phase angle =0°

We need to calculate the spring constant

Using formula of restoring force

We need to calculate the time period

Using formula of time period

Put the value into the formula

At t = 0, the maximum displacement was 5 cm

So, The equation of the time-dependent function of the position

Put the value into the formula

Hence, The equation of the time-dependent function of the position is

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

  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.