Figure 13.21(a) shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. A force $F$ applied at the free end stretches the spring. Figure 13.21(b) shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force $F$. ![Figure 13.21](/figures/oscillations/spring-clamped-wall-vs-both-ends-free-masses-fig-13-21.svg) (a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its fre

Exercises · 13.13

k

(a) What is the maximum extension of the spring in the two cases?

(b) If the mass in Fig.

(a) and the two masses in Fig.

(b) are released, what is the period of oscillation in each case?

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Concept understanding — Spring-Mass System Period

The Spring-Mass System: Why a Weight on a Spring Oscillates

Imagine hanging a weight from a spring and giving it a gentle tug downward. It bounces back up, overshoots, comes down again, and keeps going. That rhythmic up-and-down motion is simple harmonic motion, and the time it takes to complete one full bounce — down, up, and back to the start — is called the period $T$ .

The key question: what determines how fast or slow this bouncing happens? Intuition says two things matter: how heavy the weight is, and how stiff the spring is.

The Intuition

A heavier mass is harder to accelerate. It lumbers along, so the oscillation is slower — the period gets longer. A stiffer spring (one with a larger spring constant $k$ ) pulls back harder for the same stretch. That stronger restoring force whips the mass back faster, so the period gets shorter.

So the period $T$ should increase with mass $m$ and decrease with stiffness $k$ . The exact relationship turns out to be:

T = 2\pi \sqrt{\frac{m}{k}}

The $2\pi$ factor comes from the geometry of circular motion (which underlies all simple harmonic motion), and the square root tells us that doubling the mass only multiplies the period by $\sqrt{2} \approx 1.4$ , not by 2.

Where Does This Formula Come From?

For a mass on an ideal spring, the restoring force is Hooke's law: $F = -kx$ , where $x$ is the displacement from equilibrium. Newton's second law $F = ma$ gives:

m \frac{d^2x}{dt^2} = -kx

This is a differential equation whose solution is a sine or cosine wave. The angular frequency $\omega$ (how fast the oscillation goes in radians per second) comes out as:

\omega = \sqrt{\frac{k}{m}}

Since period $T$ is the time for one complete cycle, and one cycle corresponds to $2\pi$ radians, we have $T = 2\pi / \omega$ . Substituting $\omega$ gives the formula above.

Tip

If you ever forget which variable goes in the numerator, remember: mass in the numerator makes the period longer (heavier = slower), and stiffness in the denominator makes the period shorter (stiffer = faster). The square root just moderates the effect.

What the Formula Tells You

Mass and period: Double the mass → period increases by $\sqrt{2}$ (about 1.4 times). Quadruple the mass → period doubles.
Stiffness and period: Double the spring constant → period decreases by $\sqrt{2}$ (about 0.7 times). Quadruple the stiffness → period halves.
Independence from amplitude: The period does not depend on how far you pull the mass initially. A small bounce and a big bounce take exactly the same time. This is the hallmark of simple harmonic motion for an ideal spring. …

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The maximum extension is $\dfrac Fk$ in both figures -- a spring's extension depends only on the tension running through it, whether one end is clamped or both ends are pulled.
The period is $T_a=2\pi\sqrt{m/k}$ for the single mass on the clamped spring (Fig. a), and $T_b=2\pi\sqrt{m/(2k)}$ for the two equal masses on the free spring (Fig. b), because the two-body system oscillates with a reduced mass $\mu=m/2$ .

(a) Maximum extension

Figure (a) -- one end clamped. A force $F$ is applied at the free end. By Hooke's law:

F = kx_a \quad\Rightarrow\quad x_a = \frac Fk

Figure (b) -- both ends free, each pulled by $F$ . Think about the tension running through the spring. When equal and opposite forces $F$ pull on the two ends, every cross-section of the spring carries the same internal tension $F$ -- exactly the same tension as in Figure (a), where the wall simply supplies the reaction force at the clamped end. Since a spring's extension is governed purely by the tension in it:

x_b = \frac Fk

So the two cases give an identical maximum extension, $\dfrac Fk$ .

Watch out

A tempting but wrong approach equates the work done by $F$ to the spring's stored energy ( $Fx=\tfrac12kx^2$ ), giving $x=2F/k$ . That treats $F$ as if it does work through the full extension while all of it converts to spring PE -- but the equilibrium extension under a statically applied force is simply read off Hooke's law, $F=kx$ , not the energy method (which instead describes a suddenly released mass overshooting to $2F/k$ ).

(b) Period of oscillation

Figure (a) -- single mass $m$ on a spring of constant $k$ :

$T_a = 2\pi\sqrt{\frac mk}$ …

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(a) Maximum extension

(b) Period of oscillation

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