Change in time period of pendulum

[Vibhor]

Homework Statement

Q The length of a simple pendulum executing SHM is increased by 21% .The percentage increase in the time period of the pendulum of increased length is
a) 11%
b) 21%
c)42%
d)10%

Homework Equations

$T = 2\pi\sqrt{\frac{L}{g}}$

The Attempt at a Solution

$T^2 = 4{\pi}^2\frac{L}{g}$

By differentiating the above expression and dividing the resultant equation by the above equation, $\frac{2dT}{T} = \frac{dL}{L}$

$\frac{2dT}{T}$ x $100 = \frac{dL}{L}$ x $100$

$\frac{dL}{L}$ x$100 = 21$

Therefore , $\frac{dT}{T}$ x $100 = 10.5$ . But this isn't correct .

Please help me find the error in my approach .

Many Thanks

 

[Biker]

There is a really easy way to solve this question. If it increased by 21%, How can you put that in the equation.

Lets just say that I am gaining profit of 5% every month. If my original balance is 100$, and I want to know my balance after one month. Well I can just calculate
100 + 100 * 0.05
Which can be 1.05(100)

So you can do the same to the pendulum.

About your equation above how did you derive that equation? can you post your steps?

 

[ehild]

The approximation f(x+Δx)=f(x)+f '(x) Δx can be applied only when Δx/x << 1, which is not true now. Calculate the ratio between the two time periods from the original formula.

 

[Vibhor]

The approximation f(x+Δx)=f(x)+f '(x) Δx can be applied only when Δx/x << 1, which is not true now.

Great !

only when Δx/x << 1

Should that be Δx →0 ?

 

[ehild]

You know Δx, it is a fixed quantity.
In the limit, lim f(x+Δx)=f(x) when Δx→0.

 

[Vibhor]

Sorry . I did not understand your reply .

 

[ehild]

Great !
Should that be Δx →0 ?

No, Δx is a fixed quantity. Ignore my second sentence.

 

[Vibhor]

Ok . Thanks .