Grade 11Mechanics

A simple pendulum has a time period T1 when on the earths surface and T2 when taken to height R above the earths surface where R is the radius of the earth the value of T1 /T2 is

Aishwarya N

8 Years agoGrade 11

2 Answers

Arun

8 Years ago

Dear Student

T=2pi*sqrt(l/g)

g is inversly proportional to (R+h)² …..........h is height above the earth’s surface

therefore T is proportional to R+h

T₂/T₁= (R+h)/R =2............ putting h=R

Hence

T₁ / T₂ = 1/2 RegardsArun (askIITians forum expert)

bharat

8 Years ago

time period of a pendulum on earth is

T=2 $\prod$ *sqrt(r/g)

g is the acceleration due to gravity

as now the distance is r+h

since g is inversely proportional to (r+h)^2

where h is the height above the earth’s surface

hence t is proportinal to r+h

now since h=r

t2/t1=(r+h)/r=2

hence

t1/t2=1/2

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A simple pendulum has a time period T1 when on the earths surface and T2 when taken to height R above the earths surface where R is the radius of the earth the value of T1 /T2 is

Aishwarya N

2 Answers

Arun

bharat

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