12.
Note the spring-and-mass arrangement below. With no mass the spring has a rest length
x0. Now a mass m is introduced as shown. Which curve in the gray box best represents
the variation in TOTAL energy of this system?
a. Curve number 1
b. Curve number 2
c. Curve number 3
d. Curve number 4
The total mechanical energy of the spring–mass system is independent of displacement, so the correct option is curve number 3, the horizontal line.
Introduction
In a frictionless spring–mass system, the block performs simple harmonic motion (SHM) and continuously exchanges energy between kinetic and potential forms. However, the total energy graph of a spring mass system as a function of displacement stays constant because mechanical energy is conserved. This question asks which of the four given curves correctly
Physics of the Spring–Mass System
Consider a vertical spring of constant k with a mass m attached, oscillating about its equilibrium position
x0. The elastic potential energy stored in the spring at displacement x from equilibrium is
\( U = \tfrac{1}{2}k(x – x_0)^2 \), and the kinetic energy of the mass is
\( K = \tfrac{1}{2}mv^2 \). [web:22]
The total mechanical energy is
\( E = K + U = \tfrac{1}{2}mv^2 + \tfrac{1}{2}k(x – x_0)^2 \), which, for ideal SHM without friction, is a constant equal to
\( \tfrac{1}{2}kA^2 \), where A is the amplitude. [web:22][web:29]
Therefore, as the mass moves between its extreme positions, kinetic and potential energies trade off, but their sum E does not depend on the instantaneous displacement x. [web:23]
Interpreting the Four Curves
In the gray inset of the problem, energy (in joules) is plotted on the vertical axis and displacement
(x − x0) on the horizontal axis.
Curve 1 (Increasing Straight Line)
This curve suggests that total energy increases linearly with displacement magnitude. That would mean the system gains or loses mechanical energy as the block moves away from equilibrium, which contradicts conservation of energy in simple harmonic motion. Hence, curve 1 cannot represent total energy.
Curve 2 (V‑Shaped, Proportional to |x − x0|)
This curve indicates that energy is zero at equilibrium and grows linearly with distance from equilibrium, resembling neither the quadratic elastic potential energy nor a constant total energy. Since total energy is not zero at equilibrium (only kinetic or potential may be zero individually), curve 2 is also incorrect.
Curve 3 (Horizontal Straight Line)
This curve shows the same energy value for all displacements, exactly matching the prediction that
\( E = \tfrac{1}{2}kA^2 \) is constant for a given oscillation.
It correctly depicts conservation of mechanical energy for an ideal spring–mass oscillator, so curve number 3 is the right answer.
Curve 4 (Downward Sloping Line)
This curve implies that total energy decreases as the block moves away from equilibrium, which again violates conservation of energy in the absence of non‑conservative forces like friction or air resistance. Therefore, curve 4 cannot represent the total energy.
Why Curve Number 3 Is Correct
Because the only forces doing work are conservative (spring force and gravity, which are both accounted for in the effective potential), the mechanical energy of the mass–spring system remains constant throughout the motion.
Any decrease in kinetic energy is exactly balanced by an increase in potential energy, and vice versa, so a plot of total energy versus displacement must be a horizontal line, which is precisely what curve 3 shows.
