Question

The equation of a transverse wave travelling on a rope is given by y=10sin(0.01x2.00t)y=10sin(0.01x2.00t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about 63 cm/sec 75 cm/s 100 cm/sec 121 cm/sec

Answer 1

Answer: The maximum transverse speed of particle in the rope is 20 cm/s.

Explanation:

Given that,

$y = 10 \sin(0.01x-2.00t)$

We know that,

The general equation a wave

$y = A\sin(kx-\omega t)$

On comparing of general equation from given equation

amplitude A = 10

$\omega = 2.00$

We know that,

The maximum transverse speed of particle

$v = A\omega$

$v = 10\times2.00$

$v = 20\ cm/s$

Hence, The maximum transverse speed of particle in the rope is 20 cm/s.

Answer 2

Answer:

62.8 cm/s

Explanation:

If we compare this to the equation

y= A Sin( kx-wt)

y= 10 Sin(pie)(0.01x X 2.00t)

We get

A = 10

w = 2(pie)

Transverse speed of a wave is

V = Aw

= 10 x 2(pie)

= 62.8 cm/sec

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