Q4. How long must a simple pendulum be to have a period of 1 s (g = 9.8 m s)?
L = gT/4π = 9.8 × 1/(4π), about 0.248 m, a little under 25 cm.
Q5. A pendulum 1 m long hangs in a lift accelerating upwards at 2.2 m s. Find its period (g = 9.8 m s).
In the lift the bob behaves as if gravity were g + a = 12.0 m s: when the bob hangs still in the lift, the string must pull with m(g + a) to give it the lift’s acceleration (a step beyond the NCERT text). So T = 2π√(1/12.0), about 1.81 s, shorter than the 2.01 s at rest. If the lift accelerated downwards, g − a would replace g; in ideal free fall there is no gravitational restoring force and hence no finite small-oscillation period. Existing relative motion can persist; the bob need not become motionless.
Q6. The length of a pendulum is increased by 21%. By what percentage does its period change?
T ∝ √L, so the period is multiplied by √1.21 = 1.1: it increases by 10%. Halving 21% to get 10.5% is a mistake here: the rule ΔT/T ≈ ½ ΔL/L holds only for small changes.