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Question: A pulse is started at a time t = 0 along the +x direction on a long, taut string. The shape of the p...

A pulse is started at a time t = 0 along the +x direction on a long, taut string. The shape of the pulse at t = 0 is given by function f(x) with f(x)={x4+1for 4<x0x+1for 0<x<10otherwise f(x) = \begin{cases} \frac{x}{4} + 1 & \text{for } -4 < x \leq 0 \\ -x + 1 & \text{for } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases} here f and x are in centimeters. The linear mass density of the string is 50 g/m and it is under a tension of 5N.

A

The shape of the string is drawn at t = 0 then the area of the pulse enclosed by the string and the x-axis is 2.5 cm².

B

The shape of the string is drawn at t = 0 then the area of the pulse enclosed by the string and the x-axis is 5 cm².

C

The transverse velocity of the particle at x = 13 cm and t = 0.015 s will be - 250 cm/s

D

The transverse velocity of the particle at x = 13 cm and t = 0.015 s will be 250 cm/s

Answer

A, C

Explanation

Solution

  1. Calculate the area of the pulse at t=0t=0 by integrating the given function f(x)f(x) over its non-zero domain (4,1)(-4, 1). The integral is split into two parts corresponding to the piecewise definition of f(x)f(x). The area is the sum of the areas of the two triangular regions formed by the pulse shape and the x-axis.

  2. Calculate the speed of the pulse vv using the given tension TT and linear mass density μ\mu with the formula v=T/μv = \sqrt{T/\mu}. Ensure units are consistent. Convert the speed to cm/s as the position xx and function ff are given in cm.

  3. The wave function at time tt is y(x,t)=f(xvt)y(x, t) = f(x - vt). The transverse velocity of a particle is vy=ytv_y = \frac{\partial y}{\partial t}. Using the chain rule, yt=f(xvt)t(xvt)=vf(xvt)\frac{\partial y}{\partial t} = f'(x - vt) \frac{\partial}{\partial t}(x - vt) = -v f'(x - vt).

  4. Evaluate the argument u=xvtu = x - vt at the given position x=13x = 13 cm and time t=0.015t = 0.015 s.

  5. Determine the slope f(u)f'(u) of the pulse shape at the calculated position uu. This requires finding the derivative of the relevant part of the piecewise function f(x)f(x).

  6. Calculate the transverse velocity vy=vf(u)v_y = -v f'(u) using the calculated speed and slope.

  7. Compare the calculated area and transverse velocity with the values given in the options to identify the correct statements.