CBSE (Science) Class 11CBSE
Share
Notifications

View all notifications

A Transverse Harmonic Wave on a String is Described By Y(X, T) = 3.0 Sin (36 T + 0.018 X + π/4) Where X And Y Are in Cm And T In S. the Positive Direction Of X Is from Left to Right. is this a Travelling Wave Or a Stationary Wave? What Are Its Amplitude and Frequency - CBSE (Science) Class 11 - Physics

Login
Create free account


      Forgot password?

Question

A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) 

Where x and y are in cm and t in s. The positive direction of x is from left to right.

(a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

(b) What are its amplitude and frequency?

(c) What is the initial phase at the origin?

(d) What is the least distance between two successive crests in the wave?

Solution 1

(a) Yes; Speed = 20 m/s, Direction = Right to left

(b) 3 cm; 5.73 Hz

(c) `pi/4`

(d) 3.49 m

Explanation:

(a) The equation of a progressive wave travelling from right to left is given by the displacement function:

y (x, t) = a sin (ωt + kx + Φ) … (i)

The given equation is:

`y(x,t) = 3.0 sin (36t + 0.018 x + pi/4)` ...(ii)

On comparing both the equations, we find that equation (ii) represents a travelling wave, propagating from right to left.

Now, using equations (i) and (ii), we can write:

ω = 36 rad/s and k = 0.018 m–1

We know that:

`v = omega/(2pi)` and `lambda = (2pi)/k`

Also

 = νλ

`:. v = (omega/2pi) xx ((2pi)/k) = omega/k`

`= 36/0.018 = 2000 "cm/s"  = 20 "m/s"`

Hence, the speed of the given travelling wave is 20 m/s.

b) Amplitude of the given wave, a = 3 cm

Frequency of the given wave: 

`v =omega/(2pi) = 36/(2xx3.14)= 5.73 "Hz"`

(c) On comparing equations (i) and (ii), we find that the initial phase angle, `phi = pi/4`

(d) The distance between two successive crests or troughs is equal to the wavelength of the wave.

Wavelength is given by the relation: k = `(2pi)/lambda`

`:. lambda = (2pi)/k = (2xx3.14)/(0.018) = 348.89 cm = 3.49 m`

Solution 2

The given equation is y(x,t) = `3.0 sin (36t + 0.018x + pi/4)` , where x and y are in cm and t in s.

a) The equation is the equation of a travelling wave, travelling from right to left (i.e along -ve direction of x because it is an equation of the type )

`y(x,t) = A sin(omegat + kx + phi)`

Here A = 3.0 cm, `omega = 36 "rad s"^(-1)`, k = 0.018 cm and `phi = pi/4`

:. Speed of wave propagation,

`v = omega/k = (36 " rad s"^(-1))/(0.018 cm^(-1)) = (36 "rad s"^(-1))/(0.018 xx 10^2 "ms^(-1)) = 20 ms^(-1)`

b) Amplitude of wave, A = 3.0 cm  = 0.03 m

Frequency of wave `v = omega/(2pi) = 36/(2pi) = 5.7` Hz

c) Initial phase at the origin, `phi = pi/4`

d) Least distance between two successive crests in the wave

`= lambda = (2pi)/k = (2pi)/0.018`

 = 349 cm = 3.5 m

  Is there an error in this question or solution?

APPEARS IN

Solution A Transverse Harmonic Wave on a String is Described By Y(X, T) = 3.0 Sin (36 T + 0.018 X + π/4) Where X And Y Are in Cm And T In S. the Positive Direction Of X Is from Left to Right. is this a Travelling Wave Or a Stationary Wave? What Are Its Amplitude and Frequency Concept: Reflection of Transverse and Longitudinal Waves.
S
View in app×