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Ask what the tension in a pendulum string is and many students answer mg cos θ. That is true only at the two turning points. Everywhere else the string must also bend the bob's path into a circle, which takes extra force. This post derives the tension at any angle, works three problems with numbers, and shows where the common shortcuts go wrong.
Where is this in the syllabus?
The NCERT Class 11 chapter Oscillations sets up the forces in section 13.8: "The radial acceleration is provided by the net radial force T − mg cos θ, while the tangential acceleration is provided by mg sin θ." The chapter Work, Energy and Power uses the same forces in Example 5.7, where a bob is given just enough speed at the bottom to reach the top of a circle with the string slack there. The Syllabus for JEE (Main) 2026 lists "motion in a vertical circle" under Unit 4, and the NEET (UG) 2026 syllabus has the same phrase. Check the syllabus for your own year.
How do you derive the tension?
Two forces act on the bob: tension T along the string towards the support, and weight mg downward. Resolve the weight along the string (mg cos θ, pointing away from the support) and along the path (mg sin θ).
Radial equation. The net force towards the support provides the centripetal acceleration v²/l:
T − mg cos θ = mv²/l, so T = mg cos θ + mv²/l.
Energy. Released from rest at θ₀, the bob has dropped a height l(cos θ − cos θ₀) by the time it reaches angle θ. The string does no work, so v² = 2gl(cos θ − cos θ₀).
Combine them:
T = mg cos θ + 2mg(cos θ − cos θ₀) = mg(3cos θ − 2cos θ₀)
At the turning points (θ = θ₀) this gives mg cos θ₀. At the bottom (θ = 0) it gives mg(3 − 2cos θ₀), which is more than mg for any swing.
Worked example 1: tension at three angles
A 0.50 kg bob on a 0.80 m string is released from rest at 60° to the vertical. Take g = 9.8 m/s², so mg = 4.9 N.
At 60° (release): T = 4.9 × cos 60° = 4.9 × 0.5 = 2.45 N.
At 30°: v² = 2 × 9.8 × 0.80 × (0.866 − 0.5) ≈ 5.74 m²/s², so v ≈ 2.40 m/s. Then T = 4.9 × 0.866 + 0.50 × 5.74 ÷ 0.80 ≈ 4.24 + 3.59 = 7.83 N. The formula agrees: 4.9 × (3 × 0.866 − 1) ≈ 7.83 N.
At the bottom: v = √(2 × 9.8 × 0.80 × 0.5) = 2.8 m/s, and T = 4.9 + 0.50 × 2.8² ÷ 0.80 = 4.9 + 4.9 = 9.8 N, exactly 2mg for a 60° release.
Worked example 2: where does a weak string snap?
The same bob hangs from a string that breaks at 8.0 N. The maximum tension, 9.8 N at the bottom, is more than that, so the string breaks on the way down.
Set 4.9 × (3cos θ − 1) = 8.0, so cos θ = (8.0/4.9 + 1) ÷ 3 ≈ 0.878 and θ ≈ 28.7°.
Speed at that moment: v² = 2 × 9.8 × 0.80 × (0.878 − 0.5) ≈ 5.92, so v ≈ 2.43 m/s.
The velocity is along the tangent, so from that instant the bob moves as a projectile launched at 28.7° below the horizontal, towards the bottom of the swing.
Worked example 3: choosing the release angle
For which release angle is the tension at the bottom exactly twice the tension at release? Set (3 − 2cos θ₀) ÷ cos θ₀ = 2. Then 3 = 4cos θ₀, so cos θ₀ = 0.75 and θ₀ ≈ 41.4°.
What is the bob's acceleration at each point?
At the turning points v = 0, so there is no centripetal acceleration; the whole acceleration is tangential, g sin θ₀. For the 60° release that is 9.8 × 0.866 ≈ 8.49 m/s². At the bottom the tangential force is zero and the acceleration is purely centripetal, v²/l = 2g(1 − cos θ₀) = 9.8 m/s². The bob is never in equilibrium during the swing, not even at the turning points where it stops for an instant.
Which mistakes cost marks?
Writing T = mg cos θ at every angle. It holds only at the turning points.
Writing T = mg at the bottom. That is the tension of a bob hanging still.
Using the small-angle SHM speed for a 60° swing. Energy conservation is exact; the SHM result is an approximation for small angles.
Getting the radial sign wrong. The weight's radial component points away from the support, so T must be larger than it.
The same radial equation appears in our post on the conical pendulum, where the bob moves in a horizontal circle instead.
If you can derive the formula but freeze on a new set-up, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning at each step.
A practice routine
Day 1: Derive T = mg(3cos θ − 2cos θ₀) from a blank page, with a force diagram.
Day 2: Redo example 1 for a 90° release and confirm the bottom tension is 3mg.
Day 3: Two breaking-string problems and one projectile follow-up.
Day 4: NCERT Example 5.7, then find the tension at the side of the circle.
Tonight, find the tension at 45° for the 60° release in example 1. You should get about 5.49 N.
