# Change in time period of pendulum

1. Mar 4, 2016

### Vibhor

1. The problem statement, all variables and given/known data

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%

2. Relevant equations

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

3. 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 .

Many Thanks

Last edited: Mar 5, 2016
2. Mar 4, 2016

### 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.

3. Mar 5, 2016

### 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.

4. Mar 5, 2016

### Vibhor

Great !

Should that be Δx →0 ?

5. Mar 5, 2016

### ehild

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

6. Mar 5, 2016

7. Mar 5, 2016

### ehild

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

8. Mar 5, 2016

### Vibhor

Ok . Thanks .