(a) Obtain an expression for the time period T of a simple pendulum. The time period depends on : (i) mass 'm'

Question 58 of 66

Q. (a) Obtain an expression for the time period T of a simple pendulum. The time period depends on :

(i) mass 'm' of the bob

(ii) length 'l' of the pendulum and

(iii) acceleration due to gravity 'g' at the place where the pendulum is suspended. (Constant k = 2π) OR

(b) Explain in detail the Triangle Law of Vector Addition.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 5mImportance★★★★★

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Concept understanding — Simple Pendulum Period

Simple Pendulum Period: From Intuition to Formula

Imagine tying a small weight to a string, holding the other end fixed, and giving it a gentle push. It swings back and forth — that’s a simple pendulum. The question is: what determines how fast it swings? Does a heavier bob swing faster? Does a longer string make it slower?

Let’s start with what you already feel. If you hold a short string (say 20 cm) and swing it, the bob zips back and forth quickly. If you use a long string (say 1 m), the swing is noticeably slower. So length matters — longer means slower.

What about the weight? Try a light plastic bob and a heavy metal one of the same size, on the same string. You’ll find they swing at the same speed. That’s surprising — heavier things don’t fall faster, and here they don’t swing faster either. So mass does not affect the period (the time for one complete back-and-forth swing).

What about how hard you push? If you give a big push, the bob swings wider, but does it take more time? For small swings (small angles, say less than about 15°), the period is almost the same regardless of amplitude. That’s the key: for small oscillations, the pendulum is isochronous — its period is independent of amplitude.

Note

This is only true for small angles. If you pull the bob to 60° and let go, the period becomes noticeably longer. In most exam problems, you assume “small oscillations” (usually < 10°).

The Precise Statement

For a simple pendulum of length $L$ (measured from pivot to centre of bob), swinging with small amplitude in a uniform gravitational field $g$ , the time period $T$ (time for one complete oscillation) is:

T = 2\pi \sqrt{\frac{L}{g}}

That’s it. No mass term. No amplitude term (for small angles).

T = 2\pi \sqrt{\frac{L}{g}}

Why does this formula make sense?

$L$ in numerator: longer string → larger $T$ (slower swing). Doubling $L$ multiplies $T$ by $\sqrt{2} \approx 1.4$ .
$g$ in denominator: stronger gravity (larger $g$ ) → smaller $T$ (faster swing). On the Moon ( $g \approx 1.6$ m/s²), the same pendulum swings much slower.
$2\pi$ : comes from the mathematics of simple harmonic motion — the pendulum’s motion is approximately sinusoidal for small angles.

Tip

To remember: the formula is identical to that of a mass on a spring ( $T = 2\pi\sqrt{m/k}$ ), but here the “restoring force per unit displacement” is $mg/L$ , so the effective “ $k$ ” is $mg/L$ , giving $T = 2\pi\sqrt{L/g}$ .

Common exam pitfalls

Don’t confuse $L$ with amplitude. $L$ is the string length, not how far you pull it. …

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Assuming T = k m^a l^b g^c and matching dimensions on both sides gives a = 0, b = 1/2, c = -1/2, so T = 2 pi sqrt(l/g).

This question offers a choice between (a) deriving the time period of a simple pendulum by the dimensional method, and (b) the triangle law of vector addition; part (a) is answered here.

Let the time period T of a simple pendulum depend on the mass m of the bob, the length l of the pendulum, and the acceleration due to gravity g, as a power-law relation:

T = k m^a l^b g^c ... (1)

where k is a dimensionless constant (given as k = 2 pi).

Writing the dimensions of each quantity:

[T] = T^1

[m] = M^1

[l] = L^1

[g] = L^1 T^-2

Substituting into equation (1):

M^0 L^0 T^1 = M^a L^b (L T^-2)^c = M^a L^(b+c) T^(-2c)

Comparing powers of M, L, T on both sides:

Power of M: 0 = a gives a = 0

Power of T: 1 = -2c gives c = -1/2

Power of L: 0 = b + c gives b = -c = 1/2

So the relation becomes

T = k m^0 l^(1/2) g^(-1/2) = k sqrt(l/g)

…

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Q.(a) Obtain an expression for the time period T of a simple pendulum. The time period depends on :

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