A student measures the time taken for 40 complete oscillations of a simple pendulum using … — JAMB UTME

A student measures the time taken for $40$ complete oscillations of a simple pendulum using a digital stopwatch that has a negative zero error of $-0.40\text{ s}$ . If the stopwatch displays a reading of $47.60\text{ s}$ for the oscillations, what will be the correct period of oscillation when the pendulum's length is reduced to one-fourth ( $1/4$ ) of its initial length?

$0.60\text{ s}$ Answer
B
$0.59\text{ s}$
C
$0.30\text{ s}$
D
$2.40\text{ s}$

Answer

The correct period of oscillation for the shortened pendulum is

0.60\text{ s}

The correct answer is obtained by first adjusting the measured time for negative zero error (

47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}

), calculating the initial period per oscillation (

48.00 / 40 = 1.20\text{ s}

), and then taking into account that period varies with the square root of pendulum length. Reducing length to

1/4

reduces period by a factor of

\sqrt{4} = 2

, yielding

0.60\text{ s}

Step-by-Step Solution

Correct the recorded total time for instrument zero error

True total time

t = \text{Uncorrected Time} - \text{Zero Error} = 47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}

Negative zero error means the timer reads less than the actual elapsed time, so the absolute value of the error must be added back.

Calculate the initial period of oscillation

Initial period

T_1 = \frac{t}{N} = \frac{48.00\text{ s}}{40} = 1.20\text{ s}

Period is defined as the time per single oscillation.

Apply the pendulum scaling relationship for length and period

New period

T_2 = T_1 \times \sqrt{\frac{L_2}{L_1}} = 1.20\text{ s} \times \sqrt{\frac{1}{4}} = 1.20\text{ s} \times 0.5 = 0.60\text{ s}

The period of a simple pendulum is proportional to

\sqrt{L}

, so quartering the length reduces the period to half of its original value.

Key Concept

Measurement of time using a stopwatch with zero error correction combined with simple pendulum period dependence on length.

Estimated Time:2m 0s

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