Question

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is the new period? A) T B) T/4 C) 2T D) sqrt(2)T (the square root of two multiplied with T) E) 4T

Answer 1

Answer:

D) $\sqrt(2)T$

Explanation:

The period of a pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the pendulum's length

g is the acceleration due to gravity

From the formula, we observe that the period does not depend on the mass of the pendulum, but only on the length L.

In this problem, the length of the pendulum L is doubled:

L' = 2 L

So the new period of the pendulum will be

$T'=2\pi \sqrt{\frac{L'}{g}}=2\pi \sqrt{\frac{(2L)}{g}}=\sqrt{2}(2\pi \sqrt{\frac{L}{g}})=\sqrt{2} T$

So the period of the pendulum increases by a factor $\sqrt{2}$.

Answer 2

The new period is D) √2 T

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Further explanation

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension ( compression ) ( m )

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$\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}$

where:

T = period of simple pendulum ( s )

L = length of pendulum ( m )

g = gravitational acceleration ( m/s² )

Let us now tackle the problem!

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Given:

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

Asked:

final period = T₂ = ?

Solution:

$T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}$

$T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}$

$T : T_2 = \sqrt{L} : \sqrt{2L}$

$T : T_2 = 1 : \sqrt{2}$

$\boxed {T_2 = \sqrt{2}\ T}$

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Learn more

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302
• Young Modulus : brainly.com/question/9202964
• Simple Harmonic Motion : brainly.com/question/12069840

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Answer details

Grade: High School

Subject: Physics

Chapter: Elasticity