Question

A transverse wave on a string travelling along +ve x-axis has been shown in the figure below:

The mathematical form of the shown wave is $y=\left(3.0cm\right)sin[2\pi \times 0.1t-\frac{2\pi }{100}x]$ where t is in seconds and x is in centimeters. Find the total distance travelled by the particle at (1) in 10 min 15 s, measured from the
instant shown in the figure and direction of its motion at the end of this time.

• A. 6 cm, in upward direction
• B. 6 cm, in downward direction
• C. 738 cm, in upward direction
• D. 732 cm, in upward direction

Answer 1

The correct option is C 738 cm, in upward direction

At the moment shown in the figure. Particle at 1 is moving in the downward direction.
We have, $T=\frac{1}{0.1s}=10s$.
In one complete cycle, particle travels a distance, 4 times the amplitude. So, in time 10 min 15 s, i.e.,
615 s which means 61 full + 1 half cycles, the distance travelled
$=(4\times 3)\times 61+(2\times 3)\times 1=732+6=738cm$
At the start the particle is moving downwards hence after the last half cycle it is moving in the upward direction.