A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended fr

Exercises · 13.8

Q. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

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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 spring constant is found from the maximum load and extension, then the period of oscillation gives the suspended mass. The weight of the body is approximately 219.5 N.

Concept and Intuition

This problem connects two ideas: the static calibration of a spring balance and the dynamic oscillation of a mass on a spring. The balance reads 0 to 50 kg over 20 cm — that tells us the spring constant $k$ . When a mass $m$ is hung and set oscillating, the period $T = 2\pi \sqrt{m/k}$ reveals $m$ . The weight is then $mg$ .

The key insight: the same spring that stretches under a known load also governs the oscillation frequency. The static data gives $k$ ; the dynamic data gives $m$ .

Step-by-step solution

1. Find the spring constant $k$ from the calibration data.

The scale reads 0 to 50 kg, meaning a mass of 50 kg produces a full-scale deflection of 20 cm. The force at full load is:

F = mg = 50 \times 9.8 = 490 \text{ N}

Hooke’s law: $F = kx$ , where $x = 20 \text{ cm} = 0.20 \text{ m}$ .

k = \frac{F}{x} = \frac{490}{0.20} = 2450 \text{ N/m}

k = \frac{mg}{x} = \frac{50 \times 9.8}{0.20} = 2450\ \text{N/m}

2. Relate the period of oscillation to the suspended mass.

For a mass $m$ on a spring of constant $k$ , the period of simple harmonic motion is:

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

Given $T = 0.6$ s, square both sides:

T^2 = 4\pi^2 \frac{m}{k}

Solve for $m$ :

m = \frac{k T^2}{4\pi^2}

3. Substitute values to find $m$ .

$m = \frac{2450 \times (0.6)^2}{4\pi^2} = \frac{2450 \times 0.36}{4 \times 9.8696}$ …

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Q.A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Concept and Intuition

Step-by-step solution

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