Answer:
a)1.37 s
b)∞ ( Infinite)
Explanation:
Given that
L= 47 cm ( 1 m =100 cm)
L= 0.47 m
a)
On the earth :
Acceleration due to gravity = g
We know that time period of the simple pendulum given as
[tex]T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}[/tex]
Here
[tex]g_{eff}= g[/tex]
Now by putting the values
[tex]T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}[/tex]
T=1.37 s
b)
Free falling elevator :
When elevator is falling freely then
[tex]g_{eff}= 0[/tex] ( This is case of weightless motion)
Therefore
[tex]T=2\pi\sqrt{ \dfrac{L}{0}[/tex]
T=∞ (Infinite)
(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.
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