Question

If the period of a simple pendulum is T and we increase its length so that it is four times longer, what will the new period be? T/4 It is unchanged. T/2 2T 4T

Answer 1

Answer:

2T

Explanation:

The period of a simple pendulum is:

T = 2π √(L/g)

If L is increased by a factor of 4, then:

2π √(4L/g)

4π √(L/g)

2T

So the period is doubled.

Answer 2

The new period will be 2T

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Further explanation

Simple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.

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The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.

$\boxed{T = 2 \pi\sqrt{\frac{m}{k}}}$

T = Periode of Spring ( second )

m = Load Mass ( kg )

k = Spring Constant ( N / m )

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The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.

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

T = Periode of Pendulum ( second )

L = Length of Pendulum ( kg )

g = Gravitational Acceleration ( m/s² )

Let us now tackle the problem !

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Given:

initial period of a simple pendulum = T₁ = T

initial length of simple pendulum = L

final length of simple pendulum = 4L

Asked:

final period of a simple pendulum = T₂ = ?

Solution:

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

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

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

$T : T_2 = \sqrt{L} : \sqrt{4L}$

$T : T_2 = 1 : 2$

$\boxed{T_2 = 2T}$

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Learn more

• Model for Simple Harmonic Motion : https://brainly.com/question/9221526
• Force of Simple Harmonic Motion : https://brainly.com/question/3323600
• Example of Simple Harmonic Motion : https://brainly.com/question/11892568

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Answer details

Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion