Home
Class 11
PHYSICS
A transverse wave travelling on a stretc...

A transverse wave travelling on a stretched string is is represented by the equation
`y =(2)/((2x - 6.2t)^(2)) + 20`. Then ,

A

(a)velocity of the wave is `3.1m//s`

B

(b) amplitude of the wave is `0.1m`

C

(c)frequency of the wave is `20 Hz`

D

(d) wavelength of the wave is `1m`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we start with the given wave equation: \[ y = \frac{2}{(2x - 6.2t)^2} + 20 \] ### Step 1: Identify the parameters from the wave equation The standard form of a transverse wave is generally given as: \[ y = A \sin(kx - \omega t) + d \] Where: - \( A \) is the amplitude, - \( k \) is the wave number, - \( \omega \) is the angular frequency, - \( d \) is the vertical shift. In our equation, we can identify: - The term \( (2x - 6.2t) \) suggests that \( k = 2 \) and \( \omega = 6.2 \). ### Step 2: Calculate the amplitude From the equation, we can see that the amplitude \( A \) can be derived from the term in the numerator: \[ A = \frac{2}{20} = 0.1 \, \text{meters} \] ### Step 3: Calculate the velocity of the wave The velocity \( v \) of the wave can be calculated using the formula: \[ v = \frac{\omega}{k} \] Substituting the values we found: \[ v = \frac{6.2}{2} = 3.1 \, \text{meters/second} \] ### Step 4: Calculate the frequency of the wave The frequency \( f \) can be calculated using the relationship: \[ f = \frac{\omega}{2\pi} \] Substituting the value of \( \omega \): \[ f = \frac{6.2}{2\pi} \approx 0.986 \, \text{Hertz} \] ### Step 5: Calculate the wavelength of the wave The wavelength \( \lambda \) can be calculated using the formula: \[ \lambda = \frac{2\pi}{k} \] Substituting the value of \( k \): \[ \lambda = \frac{2\pi}{2} = \pi \, \text{meters} \approx 3.14 \, \text{meters} \] ### Summary of Results - Velocity of the wave \( v = 3.1 \, \text{m/s} \) - Amplitude \( A = 0.1 \, \text{m} \) - Frequency \( f \approx 0.986 \, \text{Hz} \) - Wavelength \( \lambda \approx 3.14 \, \text{m} \) ### Final Conclusion Based on the calculations, the correct values are: - Velocity: 3.1 m/s (Correct) - Amplitude: 0.1 m (Correct) - Frequency: Approximately 0.986 Hz (Not 20 Hz) - Wavelength: Approximately 3.14 m (Not 1 m)

To solve the problem, we start with the given wave equation: \[ y = \frac{2}{(2x - 6.2t)^2} + 20 \] ### Step 1: Identify the parameters from the wave equation The standard form of a transverse wave is generally given as: ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • WAVE MOTION

    DC PANDEY ENGLISH|Exercise Level 2 Subjective|7 Videos
  • WAVE MOTION

    DC PANDEY ENGLISH|Exercise Subjective Questions|2 Videos
  • WAVE MOTION

    DC PANDEY ENGLISH|Exercise Level 2 Single Correct|5 Videos
  • VECTORS

    DC PANDEY ENGLISH|Exercise Medical enrances gallery|9 Videos
  • WORK, ENERGY & POWER

    DC PANDEY ENGLISH|Exercise Level 2 Comprehension Based|2 Videos

Similar Questions

Explore conceptually related problems

A transverse wave is travelling on a string. The equation of the wave

A transverse wave propagating on a stretched string of linear density 3 xx 10^-4 kg-m^-1 is represented by the equation y=0.2 sin (1.5x+60t) Where x is in metre and t is in second. The tension in the string (in Newton) is

Knowledge Check

  • The speed of transverse wave on a stretched string is

    A
    directly proportional to the tension in the string
    B
    directly proportional to the square root of the tension
    C
    inversely proportional to live
    D
    inversely proportional to sqyare of tension
  • Similar Questions

    Explore conceptually related problems

    A transverse wave travelling on a taut string is represented by: Y=0.01 sin 2 pi(10t-x) Y and x are in meters and t in seconds. Then,

    Two wave pulses travelling along the same string are represented by the equation, y_(1)=(10)/((5x-6t)^2+4) and y_(2)=(-10)/((5x+6t-6)^2+4) (i) In which direction does each wave pulse travel (ii) At what time do the two cancel each other?

    A travelling wave in a stretched string is described by the equation y = A sin (kx - omegat) the maximum particle velocity is

    A wave pulse on a horizontal string is represented by the function y(x, t) = (5.0)/(1.0 + (x - 2t)^(2)) (CGS units) plot this function at t = 0 , 2.5 and 5.0 s .

    A string of length 2L, obeying hooke's law, is stretched so that its extension is L. the speed of the transverse wave travelling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed of transverse wave travelling on the string will be

    The equation of a transverse wave propagating through a stretched string is given by y=2 sin 2pi ((t)/(0.04) -x/(40)) where y and x are in centimeter and t in seconds. Find the velocity of the wave, maximum velocity of the string and the maximum acceleration.

    A travelling wave in a stretched string is described by the equation y=Asin(kx- omega t) . The ratio of a square of maximum particle velocity and square of wave velocity is