Q.A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according as y=kt2, where y is vertical distance covered and k=1ms−2, its time period becomes T2. Then, T12/T22 is (g=10ms−2) (IIT 2005) a) 5/6 b) 11/10 c) 6/5 d) 5/4
Imagine you are standing in a lift. When the lift is at rest or moving at constant speed, you feel your normal weight. The moment the lift accelerates upward, you feel heavier — the floor pushes harder against your feet. If it accelerates downward, you feel lighter, as if you are floating a little.
A pendulum is just a mass on a string. Its swing is governed by the restoring force that pulls it back toward the vertical. That force depends on gravity. So if the lift's motion changes how heavy the bob feels, the pendulum will swing differently.
The Core Idea
The period of a simple pendulum is
T=2πgL
where g is the acceleration due to gravity. In a lift, the bob experiences an effective gravitygeff that is the vector sum of actual gravity and the lift's acceleration (with sign). The period becomes
T=2πgeffL
The trick is to find geff for each case.
Case 1: Lift accelerating upward
When the lift accelerates upward with acceleration a, the floor pushes the bob upward harder. The bob feels heavier. The effective gravity is
geff=g+a
The pendulum swings slower (longer period) because the restoring force is stronger — it takes more time to complete one oscillation.
T=2πg+aL
Case 2: Lift accelerating downward
When the lift accelerates downward with acceleration a, the bob feels lighter. The effective gravity is
geff=g−a
The pendulum swings faster (shorter period). If a=g, the bob is in free fall — geff=0 — and the pendulum does not oscillate at all (it just floats). If a>g, the bob would hit the ceiling; that case is usually not considered in standard problems.
T=2πg−aL
Case 3: Lift moving at constant velocity (up or down)
If the lift moves at constant speed, there is no acceleration. The bob feels its normal weight. So geff=g, and the period is the same as on ground:
T=2πgL
Watch out
A common mistake is to think that velocity changes the period. It does not. Only acceleration matters. Constant velocity means no net force change — the pendulum behaves exactly as if the lift were stationary.
The Intuition in One Sentence
The pendulum's period changes because the lift's acceleration alters the apparent weight of the bob, which changes the restoring force that drives the swing. …
Step 1. The point of suspension moves as y=kt2, so its velocity is dy/dt=2kt and its acceleration is d2y/dt2=2k, a CONSTANT upward acceleration of magnitude 2k=2(1)=2ms−2.
Step 2. Since the support accelerates upward, the effective gravity increases: geff=g+2k=10+2=12ms−2. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2022Set ANNUAL1 markMCQ
Q.A simple pendulum is suspended from the roof of a school bus which moves in a horizontal direction with an acceleration 'a', then the time period is :
(a) T is proportional to sqrt(g^2 + a^2)
(b) T is proportional to 1/(g^2 + a^2)
(c) T is proportional to (g^2 + a^2)
(d) T is proportional to 1/sqrt(g^2 + a^2)
›Reveal solutionSolution
A pendulum in a horizontally accelerating vehicle feels an 'effective gravity' that is the vector sum of true gravity g and the pseudo-force per unit mass a: g_eff = √(g²+a²). Its time period is then T = 2π√(L/g_eff), so T decreases as g_eff (and hence as g²+a²) increases.
Work in the non-inertial frame of the bus. Two 'forces' act per unit mass on the bob: real gravity g, straight down; and the pseudo-force a, horizontal, pointing backward relative to the bus's forward acceleration. These two are perpendicular to each other, so they combine (vector sum) to give an effective gravitational field:
g_eff = √(g² + a²)
The pendulum still executes SHM about the new equilibrium direction (tilted, along g_eff instead of straight down), with time period given by the usual simple-pendulum formula but using g_eff in place of g:
Q.Can we use pendulum watch in an artificial satellite?
›Reveal solutionSolution
A pendulum clock cannot work in an artificial satellite because the satellite (and everything in it) is in free fall around the Earth, giving an effective acceleration due to gravity of zero, so the pendulum never swings.
The time period of a simple pendulum of length l is:
T = 2pisqrt(l/g)
where g is the local acceleration due to gravity acting on the pendulum bob.
An artificial satellite orbiting the Earth is continuously falling freely under gravity (that free fall is what keeps it in orbit). Consequently, everything inside the satellite, including a pendulum bob, is in a state of weightlessness: the effective (apparent) value of g experienced by objects inside the satellite is zero.