Question

34 A transverse wave passes through a string with the
equation y = 10 sin pi(0.02 x - 2t), where x is in
meter and 't' in second. The maximum velocity of
the particles in wave motion is-
(1) 63 m/s,
(2) 78 m/s,
(3) 100 m/s,
(4) 121 m/s.​

Answer 1

Answer:

Given:

Equation of a travelling qave is provided as follows :

$y = 10 \sin \{\pi(0.02x - 2t) \}$

To find:

Max velocity of the medium particles.

Concept:

We have to perform partial derivative of the given Equation wrt to time on order to get a velocity function of medium particles.

After that we have to find the max velocity as per the derived equation.

Calculation:

$y = 10 \sin \{\pi(0.02x - 2t) \}$

$\therefore \: v = \dfrac{ \delta y}{ \delta t}$

$= > v \: = 10 \{\dfrac{ \delta sin(0.02\pi x - 2\pi t) }{ \delta t } \}$

$= > v = (10 \times 2\pi)cos(0.02\pi x - 2\pi t)$

Now, max value of any cos function is 1,

Considering :

$\boxed{ \red{ \cos(0.02\pi x - 2\pi t) = 1}}$

So max value of v comes as :

$\therefore v = 10 \times 2\pi$

$= > v = 10 \times 2 \times 3.14$

$= > v = 10 \times 6.28$

$= > v = 62.8 \: m {s}^{ - 1}$

$= > v \: \approx63 \: m {s}^{ - 1}$

So correct option is 1)

Additional information:

1. Pls be careful about the fact that velocity of media particles is different from wave Velocity .

2. Wave velocity has to be calculated

as ω/K , the values obtained from the wave Equation.

3. Velocity of media particles have to be done in the above shown Method.

Answer 2

$\mathfrak{ \huge{ \red{ \underline{ \underline{ANSWER}}}}} \\ \\ \star \sf \: \blue{ \bold{Given}} \\ \\ \implies \sf \: wave \: equation \: y = 10 \sin(0.02 \pi \: x - 2 \pi \: t) \\ \\ \star \sf \: \blue{ \bold{To \: Find}} \\ \\ \implies \sf \: maximum \: velocity \: of \: the \: wave \: particle \: in \: wave \: motion \\ \\ \star \sf \: \blue{ \bold{Formula}} \\ \\ \implies \sf \: maximum \: velcity \: of \: particle \: in \: wave \: motion \: is \: given \: as \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \pink{ \bold{V{ \tiny{max}} = A \omega}}} \\ \\ \implies \sf \: transverse \: wave \: equation \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \pink{ \bold{y = A \sin( \omega \: t - kx)}}} \\ \\ \star \sf \: \blue{ \bold{Calculation}} \\ \\ \implies \sf \: comparing \: given \: equation \: with \: wave \: equation \\ \\ \implies \sf \: \orange{\omega = 2 \pi} \: and \: \orange{A = 10 m}\\ \\ \implies \sf \: now \: V{ \tiny{max}} = A \omega = 10 \times 2 \times \pi = 62.8 \: \frac{m}{s} \\ \\ \orange{ \huge{ \star}} \: \: \red{ \underline\blue{\boxed{ \huge \green{ \bold{ \sf{V{ \tiny{max}} = 63 \: \frac{m}{s} }}}}}}$