Which one of the following is an example of the force of gravity of the earth acting on a vibrating pendulum bob ?
- (a)Applied force
- (b)Frictional force
- (c)Restoring force
- (d)Virtual force
Correct — C, Restoring force. Pull a pendulum bob aside and its weight no longer points along the string. The weight splits into two parts: one along the string, which the tension cancels, and one along the arc, of size mg sin θ, which points back towards the lowest point. That second part is what drags the bob home every time it wanders, and a force that always acts back towards the equilibrium position is by definition the restoring force of the oscillation. It is also what makes the motion periodic at all — for small swings sin θ is nearly θ, the restoring force becomes proportional to the displacement, and the bob performs simple harmonic motion with a period of 2π times the square root of length over g.
- (a)Applied force — An applied force is the external push or pull a person or agent supplies. Here the only applied force was the one that displaced the bob at the start; once it is released, nothing outside the system is pushing it, and gravity was never an applied force to begin with.
- (b)Frictional force — Friction in a pendulum comes from air resistance and from the pivot. It opposes motion rather than displacement, which is why it steadily eats the amplitude away instead of driving the swing. Gravity is not friction — if it were, the bob would simply stop rather than swing back.
- (d)Virtual force — A virtual or pseudo force is the fictitious force introduced only when you insist on doing physics from an accelerating frame, such as the centrifugal force felt in a turning bus. Gravity is a real interaction between two masses and is present in every frame, so this label does not fit.
Every oscillation needs two ingredients — inertia, which carries the body past the equilibrium point, and a restoring force, which keeps pulling it back. In a spring the restoring force comes from the stretched coils; in a pendulum it is supplied by the component of the bob's weight along the arc, mg sin θ. Because the tension along the string is exactly cancelled by the other component of the weight, gravity is the whole of the restoring effect.
The wording of this item is doing the work. It does not ask what force gravity is, but what role gravity plays here, and 'restoring' is a role rather than a kind of force. That is why students who have memorised a list of force types — applied, frictional, normal, tension — often go looking for the wrong sort of answer. Two further points worth carrying away: the restoring force is proportional to the displacement only while the swing is small, which is why a pendulum's period is amplitude-independent only for small angles; and the mass cancels out of the equation of motion, which is why a heavy bob and a light one of the same string length keep the same time. NDA has tested both of those consequences in later papers.
- The restoring force on a simple pendulum bob displaced through an angle θ is mg sin θ, directed along the arc towards the mean position.
- The component of the weight along the string, mg cos θ, is balanced by the tension and does no restoring work.
- For small angles sin θ is close to θ, the restoring force becomes proportional to displacement, and the motion is simple harmonic.
- The period is 2π√(L/g), independent of the mass of the bob and, for small swings, of the amplitude.
- Reading the question as 'what type of force is gravity' rather than 'what role does gravity play here'.
- Confusing a pseudo force, which exists only in an accelerating frame, with a real one.
- Assuming the tension in the string is what pulls the bob back — it acts along the string and cannot.
NDA rotates between naming the restoring force, computing the time period, and asking which change to mass, length or g alters it.
Consider the following statements: A simple pendulum is set into oscillation. Then I. The acceleration is zero when the bob passes through the mean position. II. In each cycle the bob attains a given velocity twice. III. Both acceleration and velocity of the bob are zero when it reaches its extreme position during its oscillation. IV. The amplitude of oscillation of the simple pendulum decreases with time. Which of these statements are correct?
- (a) I and II
- (b) III and IV
- (c) I, II and IV
- (d) II, III and IV
Answer(c) I, II and IV
The same restoring force read through its consequences — zero at the mean position where the displacement vanishes, largest at the extremes, and slowly overcome by friction so the amplitude decays.
The time period of a 1 m long pendulum approximates to
- (a) 6 s
- (b) 4 s
- (c) 2 s
- (d) 1 s
Answer(c) 2 s
The numerical consequence of the same restoring force; the famous one-metre 'seconds pendulum' beats once each way in about a second.
A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be (in terms of period T of the first pendulum)
- (a) √2 T
- (b) 1/√2 T
- (c) 2√2 T
- (d) T
Answer(a) √2 T
Proof that gravity is the restoring agent and the bob's mass is not: weaken g and the swing slows, double the mass and nothing happens.
- practice — not a real PYQ
For a simple pendulum swinging through a small angle, the restoring force acting on the bob is proportional to
- (a)the square of the displacement
- (b)the displacement
- (c)the velocity of the bob
- (d)the mass of the bob only
Answer(b) the displacement — for small angles mg sin θ is close to mg θ, so the force varies as the displacement, which is the defining condition for simple harmonic motion.
- practice — not a real PYQ
The time period of a simple pendulum does NOT depend on
- (a)the length of the string
- (b)the acceleration due to gravity at the place
- (c)the mass of the bob
- (d)the square root of the length
Answer(c) the mass of the bob — the mass cancels out of T = 2π√(L/g).