the equation of a transverse wave travel ling on a rope is given by y=10sin3.14(0.01x-2.00t)where y and x are in cm and t is in seconds.the maximum transverse speed of the particle in the rope is about
options:
1. 62.8 cm/s
2. 75 cm/s
dy/dt = -10*2*3.14 cos****
max of cos is 1
so 20*3.14 = 62.8
y=10sin 3.14(0.01x-2.00t)=
= 10sin (π•x/100 - 2•π•t).
General form is
y= Asin (2π•x/λ - ω•t).
v =dy/dt =- A •ω cos (2•π•x/λ - ω•t).
v(max) = A •ω,where A =10 cm, ω = 2•π (rad/s),
v(max) = 20• π =62.8 cm/s
To find the maximum transverse speed of the particle in the rope, we need to differentiate the equation for y with respect to time (t), and then find the maximum value of the derivative.
Given equation: y = 10sin(3.14(0.01x - 2.00t))
Differentiating both sides with respect to t:
dy/dt = -20π cos(3.14(0.01x - 2.00t))(-2.00)
Now, let's find the maximum value of dy/dt by finding the values of x and t that maximize the cosine term.
The cosine function has a maximum value of 1 when its argument is equal to 0 or a multiple of 2π. So, we set the argument of the cosine to 0 and solve for t:
3.14(0.01x - 2.00t) = 0
Simplifying the equation, we have:
0.01x - 2.00t = 0
Solving for t:
t = 0.01x / 2.00
t = 0.005x
Substituting this value of t back into the derivative, we have:
dy/dt = -20π cos(3.14(0.01x - 2.00(0.005x)))
dy/dt = -20π cos(3.14(0.01x - 0.01x))
dy/dt = -20π cos(0)
Since cos(0) = 1, we have:
dy/dt = -20π
Therefore, the maximum transverse speed of the particle in the rope is |-20π| = 20π cm/s.
To approximate this value, we can use the approximation π ≈ 3.14:
Maximum transverse speed ≈ 20 * 3.14 cm/s ≈ 62.8 cm/s
Thus, the maximum transverse speed of the particle in the rope is approximately 62.8 cm/s. Therefore, the correct option is 1. 62.8 cm/s.