Atransverse wave on a string is described with the wave function y(x, t) = (0.59 cm)sin[(1.80 m−1)x − (6.00 s−1)t]. (a) what is the wave velocity? (enter the magnitude in m/s.) m/s (b) what is the magnitude of the maximum velocity (in m/s) of the string perpendicular to the direction of the motion? m/s

(a) wave velocity = 3.33 m/s².

(b) magnitude of the maximum velocity =  3.54 m/s²

Explanation:

The general equation of a traveling wave is given,

y = Asin(ωt - kx)(equation 1)

Where A = Amplitude of the wave (m)

            ω = Angular frequency (s⁻¹)

            k = Angular wave number (m⁻¹)

(a).

From the expression above, v = ω/k

Given :  y = (0.59)sin[(1.80)x − (6.00)t (equation 2)

Comparing Equation 1 and equation 2

1.8x = -kx

∴ k = -1.8 m⁻¹

And -6.00t = ωt

∴ ω = -6.00 s⁻¹

∴ v = - 6.00/-1.8 = 3.33 m/s².

wave velocity = 3.33 m/s².

(b).

We differentiate (equation 2) with respect to time (t) to get an expression for the transverse speed of the wave.

∴ dy/dt = d{(0.59)sin[(1.80)x − (6.00)t]/dt

dy/dt = 0.59(-6.00)cos(1.8x - 6.00t)

dy/dx = 0.59(-6.00)cos(1.8x - 6.00t)

The magnitude of the maximum velocity = The absolute value of the coefficient of the cosine function.

Vmax = 0.59 × 6.00 = 3.54 m/s²

∴ magnitude of the maximum velocity =  3.54 m/s²

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more