Question 362: Oscillations

Question

A simple pendulum has time period T
1
. The point of suspension is now moved upward according to the relation y = Kt
2
, (K = 1 m/s
2
) where y is the vertical displacement. The time period now become T
2
. The ratio of $${{T_1^2} \over {T_2^2}}$$ is (g = 10 m/s
2
)

Accepted Answer

Given data:
In the beginning, the
time period
of the simple pendulum is $ T_1 $.
The
point of suspension
moves
upward
according to
$   y = K t^2, \quad K = 1 \, 	ext{m/s}^2   $
Thus, the
acceleration of the support
is:
$   a = \frac{d^2y}{dt^2} = 2K = 2 \, 	ext{m/s}^2   $
$ g = 10 \, 	ext{m/s}^2 $
We have to find the ratio:
$ \frac{T_1^2}{T_2^2} $
Step 1: Relation between acceleration and effective gravity
When the support is
accelerated upward
with acceleration $ a $, the
effective gravity
for the pendulum becomes:
$ g' = g + a $
Here,
$ a = 2 \, 	ext{m/s}^2 $
so,
$ g' = 10 + 2 = 12 \, 	ext{m/s}^2 $
Step 2: Relation of time period with gravity
For a simple pendulum:
$ T = 2\pi \sqrt{\frac{L}{g_{	ext{eff}}}} $
Thus:
$ T_1 = 2\pi \sqrt{\frac{L}{g}} $
$ T_2 = 2\pi \sqrt{\frac{L}{g'}} $
Step 3: Ratio of squares of time periods
$ \frac{T_1^2}{T_2^2} = \frac{\frac{L}{g}}{\frac{L}{g'}} = \frac{g'}{g} $
Substitute values:
$ \frac{T_1^2}{T_2^2} = \frac{12}{10} = \frac{6}{5} $
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Final Answer:
$ \boxed{\frac{T_1^2}{T_2^2} = \frac{6}{5}} $
Option B

Answer

$${5 \over 6}$$

Answer

$${6 \over 5}$$

Answer

1

Answer

$${4 \over 5}$$

Oscillations

A simple pendulum has time period T 1 . The point of suspension is now moved upward according to the relation y = Kt 2 , (K = 1 m/s 2 ) where y is the vertical displacement. The time period now become T 2 . The ratio of ${{T_1^2} \over {T_2^2}}$ is (g = 10 m/s 2 )

Choose the correct answer: