As shown in the above figure, a cart is accelerating in the horizontal direction with a constant acceleration of 3–√g3𝑔 and a 50 cm50 cm long simple pendulum is suspended from its roof. Now, find the time period of small oscillations of this pendulum in second about its equilibrium position taking g=π2 m s−2
Question
As shown in the above figure, a cart is accelerating in the horizontal direction with a constant acceleration of 3–√g3𝑔 and a 50 cm50 cm long simple pendulum is suspended from its roof. Now, find the time period of small oscillations of this pendulum in second about its equilibrium position taking g=π2 m s−2
Solution
The problem seems to be missing the figure, but I'll try to solve it based on the given information.
The time period of a simple pendulum is given by the formula:
T = 2π√(l/g')
where:
- T is the time period,
- l is the length of the pendulum, and
- g' is the effective acceleration due to gravity.
In this case, the cart is accelerating, which changes the effective acceleration due to gravity. The effective acceleration due to gravity is given by:
g' = g + a
where:
- g is the acceleration due to gravity, and
- a is the acceleration of the cart.
Substituting the given values:
g' = π² m/s² + 3√g m/s²
We don't have a numerical value for g, so we can't simplify this further.
Substituting l = 0.5 m and g' into the formula for T:
T = 2π√(0.5/g')
This is the time period of the pendulum's oscillations. Without a numerical value for g', we can't simplify this further.
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