Question

a second's pendulum is inside a car, the car is moving in a circle with radius 10m and velocity 10m/s find the time period of small oscillations of pendulum about its equilibrium position..

Answer 1

@parth502 ,@Preetha plz help

Answer 2

the expression for time period is given by

\[T= 1/2 \pi \sqrt{l/g}\]

so l can be found out cause its a second's pendulum.. so whats special in this sum???.. since the car is moving in circles... there is not only one acceleration, there is another acceleration towards the centre.. so find that net acceleration.. and plug in that into the equation

Answer 3

it would be kind enough if u could plz compute the answer bcoz my the answer does not match with the book's answer/

Answer 4

show your calculations here!

Answer 5

What is the apparent gravity in the moving car?

Answer 6

the time period of second's pendulum is 2s so the length of the string is \[\frac{ g }{ \pi ^{2} }\]
the equilibrium position is at an angle of \[\pi/4\] with the vertical
now if we wish to calculate the time period of small oscillation let us take apparent gravity in the moving car i.e. is equal to \[g \cos \frac{ \pi }{ 4 } +10 \sin \frac{ \pi }{ 4 }\]
this value is approximately \[10\sqrt{2}\]
applying it in the formulae we get\[t=2\pi \sqrt{\frac{ g }{ \pi ^{2}10\sqrt{2} }}\]
taking g=10
we have the time period as\[\sqrt{2\sqrt{2}}\]

Answer 7

but the answer is \[\sqrt{2}\]
as given in the book