A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.
Answers
Answer ⇒ a = 0.98 m/s².
Explanation ⇒ When time period of Pendulum is 4 seconds. Then,
T =2π√(l/g)
4 = 2π√(1/g)
√(l/g) =2/π
l = 4g/π²
Now, when the car is moving with certain acceleration, then there will be effective g.
When pendulum is hanged in a car moving with an acceleration a, then it will have Pseudo force backward.
Net acceleration will be given by,
g' = √(a² + g²)
Thus, 2π√(l/g) = 3.99
2π√(4g/π²g') = 3.99
∴ 4√(g/g') = 3.99
∴ g/g' = (3.99/4)²
∴ g' = g(4/3.99)²
∴ g' = g × 1.005
∴ g² + a² = g²(1.005)²
∴ g² + a² = 1.01g²
∴ a² = 0.01g²
∴ a = 0.1 × g
∴ a = 0.98 m/s².
Hence, acceleration of car is 0.98 m/s².
Hope it helps.
Answer:
Explanation ⇒ When time period of Pendulum is 4 seconds. Then,
T =2π√(l/g)
4 = 2π√(1/g)
√(l/g) =2/π
l = 4g/π²
Now, when the car is moving with certain acceleration, then there will be effective g.
When pendulum is hanged in a car moving with an acceleration a, then it will have Pseudo force backward.
Net acceleration will be given by,
g' = √(a² + g²)
Thus, 2π√(l/g) = 3.99
2π√(4g/π²g') = 3.99
∴ 4√(g/g') = 3.99
∴ g/g' = (3.99/4)²
∴ g' = g(4/3.99)²
∴ g' = g × 1.005
∴ g² + a² = g²(1.005)²
∴ g² + a² = 1.01g²
∴ a² = 0.01g²
∴ a = 0.1 × g
∴ a = 0.98 m/s².