What is the Velocity and Acceleration of a Transverse Wave on a Stretched Cord?

[Eternal Sky]

Homework Statement

The left-hand end of a long horizontal stretched cord oscillates transversely in SHM with frequency f = 250 Hz and amplitude 2.6 cm. The cord is under a tension of 140 N and has a linear density $$\mu$$ = 0.12 kg/m. At t = 0, the end of the cord has an upward displacement 1.6 cm and is falling. What is the velocity and acceleration at t = 2.0 seconds at a point on the string that is 1.00 m from the left-hand end?

Homework Equations

$$v_{y} = -D_{M} \omega \cos (kx - \omega t)$$

$$v = \sqrt{\frac{F_{T}}{\mu}}$$

$$v = \lambda f$$

The Attempt at a Solution

I determined the wave velocity using the equation,
$$v = \sqrt{\frac{F_{T}}{\mu}} = \sqrt{\frac{140 N}{.12 kg/m}} = 34 m/s$$

I used this to find the wavelength $$\lambda$$,

$$\lambda = \frac{v}{f} = \frac{34 m/s}{250 Hz} = 0.14 m$$

So, in the equation for the velocity of the particle,

$$k = \frac{2\pi}{\lambda} = 45/m$$

$$\omega = 2\pi f = 1570/s$$

Therefore, the equation for the particle velocity should be,

$$v = -(0.026 m)(1570/s)\cos[(45/m)(1.00 m) - (1570/s)(2.0 s)]$$
$$v = 35 m/s$$

However, the book says that the answer is 41 m/s. Can someone tell me what I am doing wrong? Any help would be appreciated.

 

[Redbelly98]

Homework Equations

$$v_{y} = -D_{M} \omega \cos (kx - \omega t)$$

Since initially, at x=0, the displacement is neither zero nor at a maximum, this equation should be modified to include a phase, i.e.

cos(kx - ωt + φ)
​

Use the initial displacement and velocity at x=0 to determine φ.

Looks like you have the right idea otherwise.

 

[Eternal Sky]

Yes, adding the phase angle made everything work out correctly.

Thanks for your help!