A pendulum clock gains 5 min/day — how should length be changed

hard CBSE JEE-MAIN NEET 4 min read

Question

A pendulum clock gains 5 minutes per day. How should the length of the pendulum be changed to make the clock keep correct time?

Solution — Step by Step

A clock gains time means it runs fast — it completes more oscillations per day than required. If it gains 5 minutes in one day (86400 seconds), it ticks through 86400 seconds of time in only 86400300=8610086400 - 300 = 86100 actual seconds.

The clock’s period T1T_1 is slightly less than the correct period T0=2T_0 = 2 seconds (for a seconds pendulum). Since the clock is fast, it is completing cycles too quickly — the period is too short.

The time period of a simple pendulum is:

T=2πLgT = 2\pi\sqrt{\frac{L}{g}}

This shows TLT \propto \sqrt{L}. If we want to increase the period (slow the clock down), we need to increase the length.

This is the physical intuition: a longer pendulum swings more slowly, taking more time per oscillation, so the clock runs more slowly.

Let T0T_0 = correct period, T1T_1 = current (too short) period.

The clock gains 5 minutes in 24 hours = 86400 seconds. This means in T1T_1 seconds (one correct day), the clock shows 86400 + 300 = 86700 seconds.

So the ratio:

T1T0=8640086400+300=8640086700=864867\frac{T_1}{T_0} = \frac{86400}{86400 + 300} = \frac{86400}{86700} = \frac{864}{867}

Wait — let’s think carefully. The clock completes 86700T0\dfrac{86700}{T_0} oscillations in one day (86400 actual seconds). For correct time, it should complete 86400T0\dfrac{86400}{T_0} oscillations.

The actual ratio:

T1T0=8640086700\frac{T_1}{T_0} = \frac{86400}{86700}

Since the clock gains, it does more oscillations → shorter period → T1<T0T_1 < T_0

Using T=2πL/gT = 2\pi\sqrt{L/g}:

T1T0=L1L0L1L0=(T1T0)2=(8640086700)2\frac{T_1}{T_0} = \sqrt{\frac{L_1}{L_0}} \Rightarrow \frac{L_1}{L_0} = \left(\frac{T_1}{T_0}\right)^2 = \left(\frac{86400}{86700}\right)^2

We need the new length LnewL_{new} such that Tnew=T0T_{new} = T_0:

LnewL1=(T0T1)2=(8670086400)2\frac{L_{new}}{L_1} = \left(\frac{T_0}{T_1}\right)^2 = \left(\frac{86700}{86400}\right)^2 LnewL1=(1+30086400)21+2×30086400=1+60086400=1+0.00694\frac{L_{new}}{L_1} = \left(1 + \frac{300}{86400}\right)^2 \approx 1 + \frac{2 \times 300}{86400} = 1 + \frac{600}{86400} = 1 + 0.00694 ΔLL0.006940.69%\frac{\Delta L}{L} \approx 0.00694 \approx 0.69\%

The length should be increased by approximately 0.69% of its current length.

To slow the clock down (from gaining to correct):

  • Increase the length of the pendulum by about 0.69% of its current length.

In a typical grandfather clock, this is done by lowering the bob (the heavy weight at the bottom of the pendulum) using an adjusting screw.

Why This Works

The period of a pendulum scales as L\sqrt{L} — a longer pendulum means a longer period, meaning the clock ticks more slowly. Raising the bob shortens the effective length (faster clock); lowering the bob increases the effective length (slower clock).

The approximation (1+x)21+2x(1 + x)^2 \approx 1 + 2x for small xx is valid here since 300/864000.0035300/86400 \approx 0.0035 is small compared to 1.

Alternative Method

Use the fractional error approach. If the clock gains nn seconds per day in NN total seconds:

ΔTT=nN=30086400=1288\frac{\Delta T}{T} = \frac{n}{N} = \frac{300}{86400} = \frac{1}{288}

Since TLT \propto \sqrt{L}: ΔTT=12ΔLL\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta L}{L}

ΔLL=2×1288=11440.694%\frac{\Delta L}{L} = 2 \times \frac{1}{288} = \frac{1}{144} \approx 0.694\%

Increase length by L144\dfrac{L}{144}. Same answer.

Common Mistake

Students often conclude “decrease the length” because “the clock is going fast and more length means more distance to cover” — this reasoning is wrong. A longer pendulum does travel farther per swing, but it swings more slowly because it needs more time to travel that arc. The period formula T=2πL/gT = 2\pi\sqrt{L/g} shows clearly that increasing LL increases TT (slows the pendulum). Always trust the formula over intuition for this type of problem.

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