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Question

A traveling wave is produced on a long horizontal string by vibrating ends up and down sinusoidally. The amplitude of vibration is 1.0cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10kgm-1 and is kept under a tension of 90N.

(a) Find the speed and the wavelength of the wave.

(b) Assume that the wave moves in the position x-direction and at t=0, the end x=0 is at its positive extreme position. Write the wave equation.

(c) Find the velocity and acceleration of the particle at x=50cm at time t=10ms.


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Solution

Step 1: Given Data:

Amplitude A=0.01m

Tension T=90N

Frequency f=2002 (because in a single oscillation the displacement becomes zero twice)

=100Hz

Mass density ρ=0.1kgm-1

Step 2: Finding speed and wavelength:

Assume the velocity as v and wavelength as λ then,

We know, v=Tρ

v=900.1

v=30ms-1

Now, λ=vf

λ=30100

λ=0.3m

λ=30cm

Step 3: Writing the wave equation

General equation of wave:

y(x,t)=Asin(kx-ωϕ)

For the given case, x=0, displacement is maximum.

As we know, k=2πλ and ω=2πν.

Thus, the wave equation takes the form

y=1cmcos2π{t0.01s-x30cm}

Step 4: Finding equations for velocity and acceleration of the particle:

Velocity is given by

v=dydt

=2π0.01sin2π{x30-t0.01}

and, a=dvdt

=4π20.012cos2πx30-t0.01

Step 5: Calculating the values of velocity and acceleration:

For the values x=50cm

and time t=0.01s

v=2π0.01sin2π{5030-0.010.01}

=2π0.01sin2π{23}

=2π0.01sin4π3

=200πsinπ3

=200π×32

=5.4ms-1

Now, for acceleration

a=4π20.012cos2πx30-t0.01

=4π20.012cos2π5030-0.010.01

=4π2×104×0.5

=2×103ms-2

Hence, the velocity and acceleration are 5.4ms-1 and 2×103ms-2.


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