Simple Pendulum Length

Question

A student in a laboratory wants a simple pendulum whose period is exactly 2 seconds using g equal to about 9.8 metres per second squared, so what length is required?

RankGuru · Accepted Answer

Starting from the pendulum relation $T = 2\pi\sqrt{L/g}$ presented in NCERT Class 11, Chapter 14 (Oscillations), we solve for the length by squaring and rearranging: $L = \dfrac{g T^2}{4\pi^2}$. Inserting $T = 2$ s and $g = 9.8$ m/s$^2$ gives $L = \dfrac{9.8 	imes 4}{4\pi^2} = \dfrac{39.2}{39.48} \approx 0.99$ m. This is the familiar length of a seconds pendulum, which takes one second for each half-swing. The option 0.25 m would give a much shorter period near one second. The option 2.0 m confuses the numerical value of the period with the length. The option 3.9 m results from forgetting to divide by $4\pi^2$ and instead using $gT^2$ directly divided by four. To verify by substituting back, a length of $0.99$ m with $g = 9.8$ m/s$^2$ gives $T = 2\pi\sqrt{0.99/9.8} = 2\pi\sqrt{0.101} = 2\pi(0.318) \approx 2.0$ s, which recovers the target period and confirms the algebra. Note that the result depends on the local value of $g$, so the same 2 s beat would need a slightly different length at a location with different gravity. A plausibility check: a length just under one metre for a 2 s pendulum is a standard laboratory result, and the units of $gT^2/(4\pi^2)$, namely $(	ext{m/s}^2)(	ext{s}^2)$, reduce to metres, confirming the answer is dimensionally and physically sound.

Simple Pendulum Length

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