Physics, 03.01.2020 22:31 CheddaDsk

What is the period of a simple pendulum 47 cm long (a) on the earth, and ( b) when it is in a freely falling elevator?

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Answer from: jadecharis5339

(a) The period of a simple pendulum 47 cm long when on the earth = 1.38 seconds

(b) The period of a simple pendulum when it is in a freely falling elevator = infinity (∞)

Period: This can be defined as the time taken for an object to complete one oscillation. The s.i unit is seconds (s)

The formula for the period of a simple pendulum is

T = 2π√(L/g) Equation 1

Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity.

(a) From the question,

Given: L = 47 cm = 0.47 m,

Constant: g = 9.8 m/s², π = 22/7 ≈ 3.14

Substitute these values into equation 1

T = 2(3.14)√(0.47/9.8)

T = 6.284√(0.048)

T = 6.284(0.219)

T = 1.38 seconds

(b) When it is in a free-falling elevator,

Then g = 0 m/s²

T = 2(3.142)√(0.47/0)

T = Infinity (∞)

Therefore, The period of the simple pendulum is (a) 1.38 seconds when it is on the earth and (b) infinity (∞) when it is in a freely falling elevator.

Learn more about Period of a pendulum here: link

Answer from: manny2275

a)1.37 s

b)∞ ( Infinite)

Explanation:

Given that

L= 47 cm ( 1 m =100 cm)

L= 0.47 m

On the earth :

Acceleration due to gravity = g

We know that time period of the simple pendulum given as

$T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}$

Here

$g_{eff}= g$

Now by putting the values

$T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}$

T=1.37 s

Free falling elevator :

When elevator is falling freely then

$g_{eff}= 0$ ( This is case of weightless motion)

Therefore

$T=2\pi\sqrt{ \dfrac{L}{0}$

T=∞ (Infinite)

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What is the period of a simple pendulum 47 cm long (a) on the earth, and ( b) when it is in a freely...

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