A sinusoidal transverse wave of amplitude ym = 6.4 cm and wavelength = 5.2 cm travels on a stretched cord. Find the ratio of the maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed.
I assume you have had calculus
y=Asinwt
where A is 6.4cm
f=wavespeed/lambda
w=2PIf=2PIwavspeed/lambda
y=Asinwt
y'=particle speed=Awcoswt=A2PI (wavespeed)/lambda
the maximum occurs at coswt=1
solve for particle speed/wavespeed.
The maximum particle speed can be found using the equation:
v_max = ω * ym,
where v_max is the maximum particle speed, ω is the angular frequency, and ym is the amplitude of the wave.
The angular frequency can be calculated using the equation:
ω = 2π / T,
where T is the period of the wave. The period can be found using the equation:
T = 1 / f,
where f is the frequency of the wave. The frequency can be calculated using the equation:
f = v / λ,
where v is the wave velocity and λ is the wavelength.
Given the wavelength λ = 5.2 cm and the wave speed v, we can calculate the frequency f and the period T using the above equations.
Next, we need to find the wave speed, which is given by:
v = λ * f.
Finally, we can calculate the ratio of the maximum particle speed to the wave speed:
Ratio = v_max / v.
Let's calculate these values step-by-step.
To find the ratio of the maximum particle speed to the wave speed, we first need to determine the wave speed.
The wave speed (v) is given by the equation v = λf, where λ is the wavelength and f is the frequency. In this case, the wavelength is given as 5.2 cm.
However, we are not given the frequency directly. But we can use the relationship between wave speed and frequency to find it. The wave speed is also equal to the product of the frequency and the wavelength, so we have v = λf.
Since we know both the wavelength (5.2 cm) and the wave speed, we can rearrange the equation to solve for the frequency:
f = v / λ
Now, we can calculate the speed of the wave.
Let's assume the wave speed is v in cm/s.
v = λf
The maximum particle speed represents the maximum speed at which a single particle in the cord moves transverse to the wave. In a transverse wave, particles move up and down or side to side in a perpendicular direction to the direction of wave propagation.
In a sinusoidal wave, the maximum particle speed occurs when the wave is at its peak amplitude. The maximum particle speed (vmax) is equal to the amplitude times the angular frequency (ω):
vmax = ymω
The angular frequency (ω) can be determined using the equation ω = 2πf, where f is the frequency of the wave.
Now we can calculate vmax.
vmax = ymω
= ym * 2πf
Substituting the expression for f, we have:
vmax = ym * 2π(v/λ)
Now that we have both vmax and v, we can calculate the ratio of the maximum particle speed to the wave speed:
Ratio = vmax / v
Substituting the expressions for vmax and v, we have:
Ratio = (ym * 2π(v/λ)) / v
Simplifying the expression:
Ratio = (2π(v/λ))
Now you can substitute the values given in the problem into the equation to find the ratio of the maximum particle speed to the wave speed.