Physics · Oscillations · NEET
The time period is 2 seconds, not 1 second. This is the most common mistake. One full oscillation (mean position to one side, back to the other side, and return to mean) takes 2 seconds. The pendulum crosses the mean position once every 1 second, so it 'ticks' every second. That single-second tick is why it is named a seconds pendulum, but the full period T is 2 s.
Use T = 2*pi*root(L/g) with T = 2 s and g = 9.8 m/s^2. Then L = g*T^2/(4*pi^2) = 9.8*4/(4*9.8696) = about 0.9927 m, which is close to 1 metre. In most NEET numericals with pi^2 = 9.8 or g = pi^2, the length comes out as exactly 1 m.
To stay a seconds pendulum, T must remain 2 s. Since T depends on g, and the Moon's g is about g/6, the required length becomes L' = g_moon*T^2/(4*pi^2), which is about 1/6 of the Earth length, roughly 0.17 m. If you keep the same 1 m length on the Moon, its period becomes longer than 2 s, so it is no longer a seconds pendulum.
No. The time period T = 2*pi*root(L/g) has no mass term. Doubling or tripling the bob mass does not change the period. Only the effective length L and the local acceleration due to gravity g decide whether a pendulum is a seconds pendulum.
Check the time period. If the total time divided by the number of oscillations equals 2 s, it is a seconds pendulum. Example: 30 oscillations in 60 s gives T = 60/30 = 2 s, so it is a seconds pendulum and its length must be about 1 m.
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.
It is a simple pendulum that completes one full swing back and forth in exactly 2 seconds. It passes the middle point once every 1 second.
About 0.9927 m on Earth, which is close to 1 metre, when g = 9.8 m/s^2. In exam problems it is usually taken as 1 m.
L = g*T^2/(4*pi^2) with T = 2 s. This gives L = g/(pi^2).
Yes. Frequency f = 1/T = 1/2 = 0.5 Hz, since the period is 2 seconds.
No. Higher altitude means slightly smaller g, so the period becomes a little more than 2 s unless you shorten the length.
