Question

The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.

Answer 1

Maximum tension will be there at an place where Tension is directly equal to mg and centripetal force.

Tmax. = mg + mv²/l

Now, minimum tension will be there, Angular displacement will be maximum. Since, when θ is maximum, then Cosθ is minimum and hence, mgcosθ will be minimum. Also, when θ is maximum then, v = 0(at extreme, since it is S.H.M.)

Thus, T min. = mgcosθ

Now, According to Question,

Tmax. = 2 T min.

mg + mv²/l = 2mgCosθ

g + v²/l = 2gCosθ

Cosθ = 1/2 + v²/2lg   -----(eq.(i))

Now, we need to remove the variable v and l. (v is the velocity at mean position, and l is the length of the string.)

At mean position, Kinetic energy = 1/2 mv²

At extreme Position, P.E. = mg(h)

Now, h = l - lcosθ (You know how !)

l is the length of the string. Now, when the strings goes to an extreme position, then vertical component of its length becomes lcosθ. (refer the diagram.) Change in height is h.

Thus, Potential energy is mgl(1 - cosθ)

Now, In S.H.M. Potential energy at extreme = Kinetic energy at mean and vice versa.

Thus,

1/2 mv² = mgl(1 - cosθ)

1/2 v² = gl(1 - Cosθ)

Put this value in eq(i).

Cosθ = 1/2 + gl(1 - Cosθ)/lg

Cosθ = 1/2 + 1 - Cosθ

2 Cosθ = 3/2

∴ Cosθ = 3/4

θ = Cos⁻¹(3/4)

Hence, the angular displacement is Cos⁻¹(3/4).

Hope it helps.

Answer 2

Answer:

Tmax. = mg + mv²/l

Now, minimum tension will be there, Angular displacement will be maximum. Since, when θ is maximum, then Cosθ is minimum and hence, mgcosθ will be minimum. Also, when θ is maximum then, v = 0(at extreme, since it is S.H.M.)

Thus, T min. = mgcosθ

Now, According to Question,

Tmax. = 2 T min.

mg + mv²/l = 2mgCosθ

g + v²/l = 2gCosθ

Cosθ = 1/2 + v²/2lg   -----(eq.(i))

Now, we need to remove the variable v and l. (v is the velocity at mean position, and l is the length of the string.)

At mean position, Kinetic energy = 1/2 mv²

At extreme Position, P.E. = mg(h)

Now, h = l - lcosθ (You know how !)

l is the length of the string. Now, when the strings goes to an extreme position, then vertical component of its length becomes lcosθ. (refer the diagram.) Change in height is h.

Thus, Potential energy is mgl(1 - cosθ)

Now, In S.H.M. Potential energy at extreme = Kinetic energy at mean and vice versa.

Thus,

1/2 mv² = mgl(1 - cosθ)

1/2 v² = gl(1 - Cosθ)

Put this value in eq(i).

Cosθ = 1/2 + gl(1 - Cosθ)/lg

Cosθ = 1/2 + 1 - Cosθ

2 Cosθ = 3/2

∴ Cosθ = 3/4

θ = Cos⁻¹(3/4)