- #1

postfan

- 259

- 0

## Homework Statement

A simple pendulum experiment is constructed from a point mass m attached to a pivot by a massless rod of length

L in a constant gravitational field. The rod is released from an angle θ0 < π/2 at rest and the period of motion is

found to be T0. Ignore air resistance and friction.

19. At what angle θg during the swing is the tension in the rod the greatest?

(A) The tension is the greatest at the point θ_g = θ_0.

(B) The tension is the greatest at the point θ_g = 0.

(C) The tension is the greatest at an angle θ_g with 0 < θ_g < θ_0.

(D) The tension is constant.

(E) None of the above is true for all values of θ0 with 0 < θ0 < π/2.

20. What is the maximum value of the tension in the rod?

(A) mg

(B) 2mg

(C) mLθ0_/T_0^2

(D) mg sin θ_0

(E) mg(3 − 2 cos θ_0)

21. The experiment is repeated with a new pendulum with a rod of length 4L, using the same angle θ0, and the period of motion is found to be T. Which of the following statements is correct?

(A) T = 2T_0 regardless of the value of θ_0.

(B) T > 2T_0 with T ≈ 2T_0 if θ_0 << 1.

(C) T < 2T_0 with T ≈ 2T_0 if θ_0 << 1.

(D) T > 2T_0 for some values of θ_0 and T < 2T_0 for other values of θ_0.

(E) T_0 and T are undefined because the motion is not periodic unless θ_0 << 1.

## Homework Equations

## The Attempt at a Solution

For 19, I just drew an FBD and realized that the tension in a pendulum is just mgcos(θ) which is maximized when θ=0 so the answer is B.

For 20, when θ=0 (where the maximum tension is) from the formula the tension is just mg which is (A).

For 21, I used the formula T=2pi*sqrt(L/g) so quadrupling the length should double the period making it (A).

Are my answers (and most importantly my justifications) right?