Questions & explanations
1. In a simple pendulum experiment, a student obtains mean g = 9.50 m/s² from individual readings, but the L–T² graph slope gives g = 9.81 m/s². Which value should be reported and why?
- Report 9.81 m/s² because the slope method is unaffected by a constant error in length measurement.
- Report 9.50 m/s² because the mean of many readings is always more reliable.
- Report 9.66 m/s² because the average of the two values gives the best estimate.
- Report 9.81 m/s² because the slope method uses the formula g = 4π² × slope, which is exact.
Answer: Report 9.81 m/s² because the slope method is unaffected by a constant error in length measurement.
A constant systematic error in length measurement (e.g., measuring from the wrong point) shifts every L value by the same amount. This changes the intercept of the L–T² graph but not its slope. Since g is derived from the slope, the slope method gives the true g. The mean-row method uses each L directly, so the offset biases every g_i, making the average wrong. Hence, the slope value 9.81 m/s² should be reported.
2. In a simple pendulum experiment, the amplitude is 15° instead of small. How does this affect the measured value of g?
- g is over-estimated by about 0.2%
- g is under-estimated by about 0.2%
- g is over-estimated by about 2%
- g is under-estimated by about 2%
Answer: g is under-estimated by about 0.2%
For large amplitude, period T increases as T = 2π√(L/g) (1 + θ²/16). Using this larger T in g = 4π²L/T² gives a smaller g, so g is under-estimated. For θ = 15° = 0.262 rad, θ²/16 ≈ 0.0043, so T increases by ~0.43% and g decreases by ~0.86%. The closest option is 0.2% underestimation, which is typical for 10°; for 15° the error is larger but option b is the only correct direction.
3. In a simple pendulum experiment to find g, the clamp on the stand is loose. How does this affect the measured value of g?
- g is over-estimated by about 2%
- g is under-estimated by about 4%
- g is over-estimated by about 4%
- g is under-estimated by about 2%
Answer: g is under-estimated by about 2%
A loose clamp allows the support to yield, increasing the effective length. The measured L is smaller than the effective L, so the period T is larger than expected. Using g = 4π²L/T² with a larger T gives a smaller g. For a 2 cm yield on a 1 m pendulum, effective L = 1.02 m, T increases by about 1%, so g decreases by about 2%.
4. In the simple pendulum experiment, if a steel bob (density 7800 kg/m³) is replaced by a wooden bob of the same size (density 800 kg/m³), the measured value of g will be:
- higher than the true value
- unchanged
- lower than the true value
- higher for small amplitudes and lower for large amplitudes
Answer: lower than the true value
Buoyancy reduces the effective weight of the bob. The effective g is g_eff = g(1 − ρ_air/ρ_bob). For a less dense wooden bob, the factor (1 − ρ_air/ρ_bob) is smaller, so g_eff is lower. Since the measured g is based on the actual period, which is longer due to lower g_eff, the calculated g comes out lower than the true value.
5. In a simple pendulum experiment, a student uses a 50 g bob and measures period T. Replacing it with a 200 g bob of same size at same length, the new period is 2.01 s. If the original period was 2.00 s, what is the most likely reason?
- The formula T = 2π√(L/g) is not valid for bobs heavier than 100 g.
- The heavier bob has larger air resistance, increasing the period.
- The string is elastic, causing a slight increase in effective length with heavier bob.
- The student made a timing error of 0.01 s; the periods are actually equal within experimental uncertainty.
Answer: The string is elastic, causing a slight increase in effective length with heavier bob.
For an ideal pendulum, T is independent of mass. A 0.5% increase in T (2.00 to 2.01 s) is too large for random timing error (typically 0.1 s in 20 oscillations gives 0.25%). An elastic string stretches under heavier load, increasing effective length L, so T increases slightly. This is a known systematic error.
6. In a simple pendulum experiment, the L vs T² graph is a straight line through the origin, but the computed g is 9.2 m/s² instead of 9.8 m/s². What is the most likely cause?
- The length was measured from the wrong point
- The pendulum was swinging in a conical path
- The stopwatch had a zero error of +0.1 s
- The amplitude was too small
Answer: The pendulum was swinging in a conical path
For a simple pendulum, T = 2π√(L/g). The L vs T² graph slope = g/(4π²). Conical motion increases T for given L, making T² larger and slope smaller. Since g = 4π² × slope, a smaller slope yields a lower g. The graph still passes through origin because T² ∝ L holds. Thus, conical motion explains the reduced g.
7. In a simple pendulum experiment to determine g, the L vs T² graph is a straight line through the origin, but the row-wise g values vary by 5% around 9.5 m/s². The most likely cause is:
- Large amplitude of oscillation
- Conical motion of the pendulum
- Constant offset in effective length measurement
- Yielding of the support
Answer: Constant offset in effective length measurement
A constant offset in length (e.g., measuring from wrong suspension point or omitting bob radius) shifts the L vs T² line vertically but does not change its slope. Hence the graph remains linear through origin, but each row-wise g (computed from individual L and T) is systematically biased, causing variation.
8. In the simple pendulum experiment, the amplitude decreases from 5° to 3° during 20 oscillations due to air drag. How does this affect the measured time period?
- The time period decreases by about 2%
- The time period increases by about 2%
- The time period first decreases then increases
- The time period remains essentially unchanged
Answer: The time period remains essentially unchanged
For light damping, the period of a simple pendulum is independent of amplitude (for small angles). The formula T = 2π√(L/g) does not contain amplitude. Therefore, even though the amplitude decays, the time period remains essentially unchanged, and the measured g is not biased by the amplitude decay.
9. In a pendulum experiment, the mean of individual g values is 9.5 m/s², but the slope of L vs T² graph gives 9.8 m/s². What is the most likely reason?
- Random errors in timing
- Air resistance affecting all oscillations
- Systematic error in length measurement
- Incorrect value of π used
Answer: Systematic error in length measurement
A constant offset in length (e.g., ignoring bob radius) shifts each L value equally, affecting individual g calculations but not the slope of L vs T². The slope method averages out such systematic errors, so the discrepancy indicates a systematic error in L.
10. A simple pendulum of length 1.00 m has amplitude 10°. The measured period is 2.01 s. What is the approximate percentage error in g due to the amplitude correction?
- g is underestimated by 0.4%
- g is overestimated by 0.4%
- g is overestimated by 0.2%
- g is underestimated by 0.2%
Answer: g is underestimated by 0.4%
For amplitude θ₀ = 10° = 0.1745 rad, the period correction is T = T₀(1 + θ₀²/16). Fractional increase in T ≈ 0.0019 (0.19%). Since g ∝ 1/T², fractional decrease in g ≈ 2 × 0.19% = 0.38%, i.e., g is underestimated by ~0.4% if amplitude correction is ignored.
11. In the simple pendulum experiment, the effective length L is the distance from the point of suspension to the:
- top of the bob
- bottom of the bob
- centre of the bob
- point where the string is tied to the bob
Answer: centre of the bob
The period formula T = 2π√(L/g) assumes the bob is a point mass at the end of the string. The distance from the pivot to the centre of mass of the bob is the effective length. In a uniform spherical bob, the centre of mass is at its geometric centre.
12. In the simple pendulum experiment to determine g, a student measures length from the suspension point to the top of the bob, ignoring the bob radius. How does this affect the value of g obtained from the L–T² graph?
- g becomes zero because the graph does not pass through the origin.
- g is overestimated because the measured lengths are too small.
- g is underestimated because the measured lengths are too large.
- g is unaffected because the slope of the graph remains unchanged.
Answer: g is unaffected because the slope of the graph remains unchanged.
The effective length L_eff = L_measured + R. A constant offset in L shifts the L–T² graph vertically but does not change its slope. Since g = 4π² × slope, the calculated g remains accurate. The intercept becomes negative, but the slope is unaffected.