Question

A wave in a string has an amplitude of $2$ cm. The wave travels in the positive direction of the x-axis with a speed of $128m{s^{ - 1}}$and it is noted that $5$ complete waves fit in $4$ m length of the string. The equation describing the wave is
(A) $y = (0.02)m\sin (7.85x + 1005t)$
(B) $y = (0.02)m\sin (15.7x - 20105t)$
(C) $y = (0.02)m\sin (15.7x + 20105t)$
(D) $y = (0.02)m\sin (7.85x - 1005t)$

Answer 1

Hint : First of all note down all the given measures and convert all the units in the same system of units. Then find the correlation between the known and the unknown terms and place the values and simplify for the required value.

Complete Step By Step Answer:
Given that:
Amplitude, A $= 2{\text{ cm = 0}}{\text{.02 m}}$
Speed, $v = 128m/s$
$5$ complete waves fit in $4$ m length of the string
$5\lambda = 4$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$\lambda = \dfrac{4}{5}$
Simplify the above expression –
$\lambda = 0.8m$
Now, using the formula $v = \lambda f$
Make required term the subject –
$f = \dfrac{v}{\lambda }$
Place the values in the above equation –
$f = \dfrac{{128}}{{0.8}}$
Simplify the above equation –
$f = 160Hz$
Angular frequency, $\omega = 2\pi f$
Place the value in the above equation-
$\omega = 2\pi \times 160$
Simplify the above equation –
$\omega = 1005$ Hz
Now, $k = \dfrac{{2\pi }}{\lambda }$
Place the values in the above equation –
$k = \dfrac{{2\pi }}{{0.8}}$
Simplify the above equation –
$\Rightarrow k = 7.85$
Hence, the required solution is $y = (0.02)m\sin (7.85x - 1005t)$
From the given multiple choices- the option D is the correct answer.

Note :
Be good in simplification of the basic mathematical expressions. Be very careful while placing the values in the equation. All the units should be in the same format either MKS (meter kilogram system) or the CGS (centi-meter gram second) system. Remember the correct formula or the conversion relation among the units to solve these types of solutions. Rest goes well, just substitution and simplification. Also, refer to the types of systems of units to know the correlation among the terms. Know the basic conversational relation and apply accordingly as per the requirement. Since one kilogram and one gram differs a lot.