You visited us 1 times! Enjoying our articles? Unlock Full Access!

Mechanical & Non-Mechanical Waves

write a gener...

Question

write a general expression for the wave function

y = (15 c m) sin [(16 π s^{- 1}) t - (\frac{π r a d}{20 c m}) x + \frac{π}{2}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y = (30 c m) sin [(16 π s^{- 1}) t - (\frac{π r a d}{20 c m}) x + \frac{π}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y = (15 c m) sin [(8 π s^{- 1}) t - (\frac{π r a d}{20 c m}) x + \frac{π}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y = (15 c m) sin [(16 π s^{- 1}) t - (\frac{π r a d}{10 c m}) x + \frac{π}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $y = (15 c m) sin [(16 π s^{- 1}) t - (\frac{π r a d}{20 c m}) x + \frac{π}{2}]$
It is given that A = 15 cm and also y = 15 cm at x = 0 and t = 0
then using y = A $sin (ω t - k x + ϕ)$
15 = 15 $sin ϕ$ $\Rightarrow$ $sin ϕ$ = 1
or $ϕ = \frac{π}{2}$ rad
Therefore the wave function is
$y = A sin (ω t - k x + \frac{π}{2})$
= $(15 c m) sin [(16 π s^{- 1}) t - (\frac{π r a d}{20 c m}) \times + \frac{π}{2}]$

Suggest Corrections

Similar questions

π rad/s

θ = \frac{π}{2} rad

x = {sin}^{- 1} 2 t \sqrt{1 - t^{2}}

y = \frac{π}{2} - {cos}^{- 1} t,

\frac{d^{2} y}{d x^{2}}

t = \frac{π}{3} .

t = 0

P

20 \sqrt{3} cm/s

P

60^{\circ}

x

α = \frac{θ}{t^{2}}

r a d / s^{2}

(θ)

600 °

10

If x = \sqrt{2^{c o s e c^{- 1} t}} and y = \sqrt{2^{{sec}^{- 1} t}} (| t | \geq 1), then \frac{d y}{d x} is equal to :

Join BYJU'S Learning Program