Experiment to determine the time period of a pendulum

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Time period of a simple pendulum is given by 2π√(L/g) .

By uniform mass distribution , I am assuming density of mass of bob to be constant .

I don't understand how time period depends on the mass of the bob of a simple pendulum .

 

[phoenix95]

Time period of a simple pendulum is given by 2π√(L/g) .

By uniform mass distribution , I am assuming density of mass of bob to be constant .

Yes, that is correct. As you have pointed out, the period of the pendulum doesn't depend on the mass.

When we say a mass attached to the end of the string of length L what we really mean is a 'point mass'. In other words the total mass is centred at that point(a point at exactly at one end of the string of length L). When you have geometrical(like sphere) objects having mass(And therefore a point mass) at the end of string length L, that changes things. Do you see how?

 

[Jahnavi]

Sorry . I missed that .

T₁ = 2π√(L+r₁)/g)

T₂ = 2π√(L+r₂)/g)

∆T = (2π/√g)(1/2√L)∆L

Since ∆L = ∆r

∆T = (2π/√g)(1/(2√L))∆r

Putting the values , ∆r = 5×10^-4 m = 0.05cm

This is not an option .

 

[phoenix95]

∆T = (2π/√g)(1/2√L)∆L

May I know how you got this equation?

 

[Jahnavi]

May I know how you got this equation?

Taking derivative of both sides of T=2π√(L/g)

 

[phoenix95]

The algebra of this is a little complicated. Right now I am not sure of an exact method.

Taking derivative of both sides of T=2π√(L/g)

and hence

∆T = (2π/√g)(1/2√L)∆L

This equation assumes infinitesimal change, not sure if it would work.

Try binomial expansion(Expand the square root).

 

[Jahnavi]

Try binomial expansion(Expand the square root).

I think it is one and the same thing . I tried it and it gives same result .

Do you get one of the options ?

 

[phoenix95]

I think it is one and the same thing . I tried it and it gives same result .

I am getting a different one. If you could show me your calculations, maybe we'll know what's wrong.

Do you get one of the options ?

I think I do.

 

[phyzguy]

Maybe I missed something, but I think you did the problem correctly. I suggest you talk to your teacher and tell them the problem is wrong.

 

[Jahnavi]

Maybe I missed something, but I think you did the problem correctly. I suggest you talk to your teacher and tell them the problem is wrong.

The problem is from a past entrance exam .It is most likely correct .

On reading the question again , I see the use of term " relative difference " . May be this is the part I am missing . But what is a relative difference ?

If I take this as ∆T/T , I do get one of the options . But then relative difference has units of time which means it is actually ∆T .

Please see if I am committing some mistake .

 

[phyzguy]

The problem is from a past entrance exam .It is most likely correct .

On reading the question again , I see the use of term " relative difference " . May be this is the part I am missing . But what is a relative difference ?

If I take this as ∆T/T , I do get one of the options . But then relative difference has units of time which means it is actually ∆T .

Please see if I am committing some mistake .

If you interpret the relative difference as ΔT/T, then your explanation makes sense. But then the relative difference would be 5E-4 (unitless), not 5E-4 seconds. In either case the problem is stated incorrectly.

 

[Jahnavi]

Maybe I missed something, but I think you did the problem correctly.

Do you feel 0.05 cm is correct answer ?

I suggest you talk to your teacher and tell them the problem is wrong.

In either case the problem is stated incorrectly.

Is problem statement wrong or the problem doesn't contain correct option i.e 0.05 cm ?

 

[phyzguy]

It depends where the error in the problem lies. If the problem means "The relative difference in periods is found to be 5E-4", then the right answer is 0.1cm. If the problem means "The difference in periods is found to be 5E-4 s", then the right answer is 0.05cm. You'd have to ask the person who wrote the problem what they meant.