Suppose a 200g mass (0.20kg) is oscillating at end of a spring upon horizontal s

Question

Suppose a 200g mass (0.20kg) is oscillating at end of a spring upon horizontal surface essentially friction free. The spring can be both stretched and compressed and has a spring constant of 360N/m. It was originally stretched a distance of 14 cm (0.14m) from its equilibrium (unstretched) position prior to release.

(a) What is initial potential energy?

(b) What is maximum velocity that the mass will reach in its oscillation? Where in the motion is this max reached?

(c) Ignoring friction, what are the values of the potential energy, kinetic energy and velocity of the mass when the mass is 7cm from the equilibrium position?

(d) How does the value of velociy computed in part c compare to that computed in part b? (ratio of values)

Explanation / Answer

Solution: Mass attached to a spring performs the linear simple harmonic motion (SHM).

Mass of the object M = 0.20 kg

Spring constant of the spring k = 360 N/m

Original stretching of the spring from the equilibrium position x = 0.14 m

Part (a) The initial elastic potential energy stored in the spring is given by,

Ui = (1/2)*k*x2

U­i = (1/2)*(360N/m)*(0.14m)2

Ui = 3.528 J

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Part (b) The angular frequency of SHM is given by,

= [k/M]

= [(360N/m)/(0.20kg)]

= 42.4264 rad/sec

The maximum velocity vmax of the mass is given by the velocity amplitude *xmax

Here maximum stretching of the spring equals to the amplitude of the oscillation, x = xmax = 0.14 m,

Thus the maximum velocity is,

vmax = *xmax

vmax = (42.4264 rad/s)*(0.14m)

vmax = 5.94 m/s

Thus the maximum velocity of the mass is 5.94 m/s and maximum velocity is reached while the mass crosses the equilibrium position.

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Part (c) Since there is no friction involved, the total mechanical energy is constant and it is T = K + U

We note from part (a) that at the extreme position, mass has only potential energy and kinetic energy is zero (as mass stops momentarily at the extreme position during oscillation).

Thus total energy

T = Ui + Ki

T = 3.528 J + 0 J

T = 3.528 J

The total mechanical energy of the system does not change during the motion.

Now at the position of x = 7cm = 0.07 m, new potential energy is given by,

U = (1/2)*k*x2

U­ = (1/2)*(360N/m)*(0.07m)2

U = 0.882 J

Thus the new kinetic energy K is given by

T = K + U

K = T – U

K = 3.528 J – 0.882 J

K = 2.646 J

And the velocity of the mass is given by,

K = (1/2)*M*v2

v2 = 2*K/M

v2 = 2*(2.646J)/(0.20kg)

v2 = 26.46

v = + 5.14 m/s and/or -5.14 m/s (as the square root of a number can two values, one positive and other negative)

When the mass is on the left side of the equilibrium position; it crosses a point of 7 cm two times; once while going towards negative extreme and other is when it coming towards the equilibrium position. Thus velocity is -5.14 m/s and +5.14 m/s respectively.

Similarly on the right side of the equilibrium position, when it going towards the extreme, velocity is +5.14 m/s and when it is returing towards equilibrium postion, velocity is -5.14m/s. We can only consider magnitude of the velocity, 5.14 m/s

The answers are, 0.882 J, 2.646 J and 5.14 m/s respectively.

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Part (d) The ratio of vmax/v = (5.94 m/s)/(5.14m/s)

vmax/v = 1.155

Thus we see that velocity is maximum while it crosses the equilibrium and it is less everywhere else.

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