# The plot of a linear relation given an equation

## Homework Statement

$$\omega (q)= \sqrt{( \frac{4f}{m})} sin\frac{qa}{2}$$  N/A

## The Attempt at a Solution

I dont understand the linear line given on the graph. For low q (or as q tends to zero) it says the relationship is linear. But as q tends to zero for the given equation I dont see how the equation goes to:

$\omega (q)= qa \sqrt{\frac{f}{m}}$

please could someone explain this, it's not a homework question, just going through the notes we have been given.

thanks in advance for any help

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ehild
Homework Helper
When the argument of the function sin(x) is small, x<<1, sin(x)≈x. So $ω(q)=\sqrt{\frac{4f}{m}} \sin(\frac{qa}{2}) = \sqrt{\frac{4f}{m}} \frac{qa}{2}$.
Pull out 4 from the square root, $ω(q)=2\sqrt{\frac{f}{m}} \frac{qa}{2}$, simplify by 2.

• rwooduk
When the argument of the function sin(x) is small, x<<1, sin(x)≈x. So $ω(q)=\sqrt{\frac{4f}{m}} \sin(\frac{qa}{2}) = \sqrt{\frac{4f}{m}} \frac{qa}{2}$.
Pull out 4 from the square root, $ω(q)=2\sqrt{\frac{f}{m}} \frac{qa}{2}$, simplify by 2.
cant believe i couldnt see that! thanks very much for your help!