Question

An astronaut notices that a pendulum which took 2.45 s for a complete cycle of swing when the rocket was waiting on the launch pad takes 1.25 s for the same cycle of swing during liftoff.

What is the acceleration (m/s²) of the rocket?(Hint: Inside the rocket, it appears that g has increased.)

Answer 1

2.84 g's with the remaining 1 g coming from gravity (3.84 g's)

Step-by-step explanation:

period of oscillation while waiting (T1) = 2.45 s

period of oscillation at liftoff (T2) = 1.25 s

period of a pendulum (T) =2π.
`\sqrt{(L)/(a) }`

where

• L = length
• a = acceleration

therefore the ration of the periods while on ground and at take off will be

`(T1)/(T2)` =(2π
`\sqrt{(L)/(a1) }` ) / (2π
`\sqrt{(L)/(a2) }`)

where

• a1 = acceleration on ground while waiting
• a2 = acceleration during liftoff

`(T1)/(T2)` =
`\frac{\sqrt{(L)/(a1) }}{\sqrt{(L)/(a2) }}`

squaring both sides we have

`((T1)/(T2))^(2)` =
`((L)/(a1) )/((L)/(a2) )`

`((T1)/(T2))^(2)` =
`(a2)/(a1)`

assuming that the acceleration on ground a1 = 9.8 m/s^{2}

`((T1)/(T2))^(2)` =
`(a2)/(9.8)`

a2 = 9.8 x
`((T1)/(T2))^(2)`

substituting the values of T1 and T2 into the above we have

a2 = 9.8 x
`((2.45)/(1.25))^(2)`

a2 = 9.8 x 3.84

take note that 1 g = 9.8 m/s^{2} therefore the above becomes

a2 = 3.84 g's

Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.