Ncert Exercises & Solutions

15.8 A transverse harmonic wave on a string is described by $y (x, t) = 3.0 \sin (36 t + 0.018 x + \frac{π}{4})$ where x and y are in cm and t in sec. 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?

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

Explanation :

The equation of a progressive wave travelling from right to left

is given by the displacement function :

y (x, t) = asin (ωt + kx + ϕ) . . . . (i)

The given equation is :

y (x, t) = 3.0 \sin (36 t + 0.018 x + \frac{π}{4}) . . . . (ii)

On comparing both the equations, we find that the equation (ii)

represents a travelling wave, propagating from right to left .

Now, using equations (i) and (ii),

ω = 36 rad / s and k = 0.018 m^{- 1}

And,

ν = \frac{ω}{2 π} and λ = \frac{2 π}{k}

So the speed of the wave : v = νλ

\begin{aligned} ∴ v & = (\frac{ω}{2 π}) \times (\frac{2 π}{k}) = \frac{ω}{k} \\ = \frac{36}{0 .018} = 2000 cm / s = 20 m / s \end{aligned}

Amplitude of the given wave, a = 3 cm

Frequency of the given wave :

ν = \frac{ω}{2 π} = \frac{36}{2 \times 3.14} = 5.73 Hz

On comparing equations (i) and (ii), we find that the initial phase angle, ϕ = \frac{π}{4} .

The distance between two successive crests

or troughs is equal to the wavelength of the wave .

Wavelength of the wave is given by :

λ = \frac{2 π}{k} = \frac{2 \times 3 .14}{0 .018} = 348 .89 cm = 3 .49 m

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