A pendulum has an angular amplitude θ. The tension in the string at extreme position is T1 and at the bottom is T2. If it is known that T2=2 T1, then what is the value of 100 cos θ ?

Question

A pendulum has an angular amplitude θ. The tension in the string at extreme position is T1 and at the bottom is T2. If it is known that T2=2 T1, then what is the value of 100 cos θ ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T1 = mg cosθ .....(1) T 2- mg =( mv 2/l)=( m /l)(2 gh ) =(2 mg /l)(1- cos θ) l =2 mg (1- cos θ) l or T 2= mg +2 mg (1- cos θ)...(2) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-55hwjhyucddaegte.jpg" /> Given T2 = 2T1 or textmg + 2 textmg (1 - textcos θ ) = text2mg cos θ or textcos θ = (3/4) = 0.75

Q. A pendulum has an angular amplitude $θ$ . The tension in the string at extreme position is $T_{1}$ and at the bottom is $T_{2}$ . If it is known that $T_{2} = 2 T_{1}$ , then what is the value of $100 cos θ$ ?

$T_{1} = m g cos θ .....$ (1)
$T_{2} - m g = \frac{m v ^{2}}{l} = \frac{m}{l} (2 g h)$
$= \frac{2 m g}{l} (1 - cos θ) l$
$= 2 m g (1 - cos θ) l$
or $T_{2} = m g + 2 m g (1 - cos θ) ... (2)$

Given $T_{2} = 2 T_{1}$
or $mg + 2 mg (1 - cos θ) = 2mg cos θ$
or $cos θ = \frac{3}{4}$
$= 0.75$

Q. A pendulum has an angular amplitude θ. The tension in the string at extreme position is T1​ and at the bottom is T2​. If it is known that T2​=2T1​, then what is the value of 100cosθ ?

T1​=mgcosθ.....(1) T2​−mg=lmv2​=lm​(2gh) =l2mg​(1−cosθ)l =2mg(1−cosθ)l or T2​=mg+2mg(1−cosθ)...(2) Given T2​=2T1​ or mg+2mg(1−cosθ)=2mg cosθ or cosθ=43​ =0.75

Q. A pendulum has an angular amplitude $θ$ . The tension in the string at extreme position is $T_{1}$ and at the bottom is $T_{2}$ . If it is known that $T_{2} = 2 T_{1}$ , then what is the value of $100 cos θ$ ?

$T_{1} = m g cos θ .....$ (1)
$T_{2} - m g = \frac{m v ^{2}}{l} = \frac{m}{l} (2 g h)$
$= \frac{2 m g}{l} (1 - cos θ) l$
$= 2 m g (1 - cos θ) l$
or $T_{2} = m g + 2 m g (1 - cos θ) ... (2)$

Given $T_{2} = 2 T_{1}$
or $mg + 2 mg (1 - cos θ) = 2mg cos θ$
or $cos θ = \frac{3}{4}$
$= 0.75$