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Andrei [34K]
3 years ago
9

A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath

er standing next to the swing. At the same time, a spaceship hurtles by at a speed close to the speed of light. According to special relativity (and ignoring the Doppler effect for this question), a person on the spaceship finds that the time for one full swing is:
A - less than 4 seconds when the spaceship is approaching the swing and longer than 4 seconds when it is moving away.
B - less than 4 seconds.
C - more than 4 seconds.
D - equal to 4 seconds.
Physics
1 answer:
Sati [7]3 years ago
4 0

Answer:

A

Explanation:

Solution:-

- According to the law of relativity the relative speed between two moving objects is inversely proportional to the the time taken.

- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

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Explanation:

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to find out

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solution

first we consider the polar coordinate (a,θ)

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I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

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and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

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