In the wave equation

y = 0.5 sin  (400t - x)m

the velocity of the wave will be:

This question was previously asked in
AIIMS BSc NURSING 2024 Memory-Based Paper
View all AIIMS BSc Nursing Papers >
  1. 200 m/s
  2. 200√2 m/s
  3. 400 m/s
  4. 400√2 m/s

Answer (Detailed Solution Below)

Option 3 : 400 m/s
Free
AIIMS BSc NURSING 2024 Memory-Based Paper
100 Qs. 100 Marks 120 Mins

Detailed Solution

Download Solution PDF

Concept:

Wave Equation:

The general form of a wave equation is y = A sin(kx - ωt), where:

A = Amplitude of the wave

k = Wave number (k = 2π / λ)

ω = Angular frequency (ω = 2πf)

t = Time

x = Position

The wave velocity v can be calculated using the relation:

v = ω / k

Calculation:

Given the wave equation: y = 0.5 sin(2π / λ (400t - x)) m, we can identify the following:

ω = 2π × 400 = 800π rad/s

k = 2π / λ

The velocity of the wave is given by:

v = ω / k = (800π) / (2π / λ) = 400λ

Since the equation is in the standard wave form, we can conclude that the velocity of the wave is 400 m/s.

∴ The velocity of the wave is 400 m/s, which corresponds to Option 3.

Latest AIIMS BSc Nursing Updates

Last updated on Jun 17, 2025

-> The AIIMS BSc Nursing (Hons) Result 2025 has been announced.

->The AIIMS BSc Nursing (Hons) Exam was held on 1st June 2025

-> The exam for BSc Nursing (Post Basic) will be held on 21st June 2025.

-> AIIMS BSc Nursing Notification 2025 was released for admission to the 2025 academic session.

More Wave Propagation in Lossy Medium Questions

Hot Links: teen patti party mpl teen patti teen patti apk download