# amplitude of guitar string

• Oct 7th 2009, 05:00 PM
IrrationalPI3
amplitude of guitar string
Hey everyone, i need a hand with a question i have for an assignment...

a guitar string is vibrating in its fundamental mode, with a node at both ends. Given the length (0.390m). The max transverse acceleration at a point in the middle of the string (8900m/s^2) and max transverse velocity of 4.00 m/s

how do i find the amplitude?

i know the equation of a standing wave is y(x,t)= Asin(kx)sin(wt)
where a is the amplitude of the standing wave and w denotes omega

in the fundamental mode the wavelength is 2L...

but im not sure where to start... can anyone point me in the right direction?
• Oct 8th 2009, 05:40 AM
Simple Harmonic Motion
Hello IrrationalPI3
Quote:

Originally Posted by IrrationalPI3
Hey everyone, i need a hand with a question i have for an assignment...

a guitar string is vibrating in its fundamental mode, with a node at both ends. Given the length (0.390m). The max transverse acceleration at a point in the middle of the string (8900m/s^2) and max transverse velocity of 4.00 m/s

how do i find the amplitude?

i know the equation of a standing wave is y(x,t)= Asin(kx)sin(wt)
where a is the amplitude of the standing wave and w denotes omega

in the fundamental mode the wavelength is 2L...

but im not sure where to start... can anyone point me in the right direction?

The centre point of the string will be oscillating with Simple Harmonic Motion, with equation of the form:

$\displaystyle x = a\sin(\omega t + \alpha)$

where $\displaystyle x$ is the displacment from the equilibrium position, $\displaystyle a$ is the amplitude and $\displaystyle \omega$ is a constant.

If we differentiate this equation we get

$\displaystyle v = \dot{x}=a\omega\cos(\omega t + \alpha)$

which gives the maximum velocity:

• $\displaystyle \text{vel}_{max}=a\omega$

Differentiating again, and substituting back from the original equation we get the characteristic equation of SHM:

$\displaystyle \ddot{x}=-\omega^2x$

So the maximum acceleration occurs when $\displaystyle x = \pm a$, and is therefore numerically

• $\displaystyle \text{acc}_{max}= a\omega^2$

Plug the values given in the question into these two expressions and eliminate $\displaystyle \omega$ to find $\displaystyle a$.