Physics · Oscillations · NEET
When the bob is pulled to a small angle theta, the weight mg splits into two parts. The part along the string (mg cos theta) is balanced by tension. The part perpendicular to the string, mg sin theta, is the restoring force that pulls the bob back to the mean position. So restoring force F = -mg sin theta. For a small angle, sin theta is nearly equal to theta (in radians), and theta = x/L where x is the displacement along the arc. So F = -mg (x/L) = -(mg/L) x. This is exactly the SHM form F = -k x with k = mg/L. Time period T = 2 pi root(m/k) = 2 pi root(m / (mg/L)) = 2 pi root(L/g). The mass m cancels out.
In the derivation the restoring force has m in it (F = -(mg/L) x), and the inertia term m/k also has m. When you divide, both m values cancel: m / (mg/L) = L/g. So the period depends only on length L and gravity g, never on how heavy the bob is. A steel bob and a plastic bob of the same string length take the same time to swing.
The real restoring force is mg sin theta, which is not directly proportional to displacement, so pure SHM would not hold. But when the angle is small (usually less than about 10 degrees), sin theta is almost equal to theta in radians. This makes the force proportional to displacement (F = -(mg/L) x), which is the condition for SHM. So T = 2 pi root(L/g) is valid only for small oscillations.
No, for small angles the amplitude does not change T. This is called isochronism. Pull the bob a little more or a little less and it still takes the same time for one swing, as long as the angle stays small. The period depends only on L and g.
L is the distance from the point of suspension to the center of mass of the bob. It is not just the string length. If the bob is a ball of radius r hung by a string of length l, the effective length is L = l + r. Using only the string length is a common mistake.
A pendulum hung from the roof of a tall building moves freely to and fro as a simple harmonic oscillator. The acceleration of the bob is 20 m/s^2 at a distance of 5 m from the mean position. The time period of oscillation is
If the mass of the bob of a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is (x/2) times its original time period. The value of x is
Savitha, a Class XI student, performs an experiment to find the effective length L of a simple pendulum. She records the time for 30 oscillations as 60 s. The length she calculates is (take pi^2 = 9.8 and g = 9.8 m/s^2)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2 pi root(L/g), where L is the effective length and g is the acceleration due to gravity. It is valid for small angle oscillations only.
No. The mass of the bob cancels out during the derivation, so the period depends only on length L and gravity g.
For small angles the restoring force F = -(mg/L) x is proportional to displacement and directed toward the mean position. This is the condition for SHM, so the pendulum behaves like an SHM oscillator.
T is proportional to root L. If L becomes 4 times, root L becomes 2 times, so T doubles.
A seconds pendulum has a time period of 2 s (1 s for each swing). Its length on Earth is about 1 m for g = 9.8 m/s^2.
