Ncert Exercises & Solutions

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 is in sec. The positive direction of x is from left to right.

(a) The wave is travelling from right to left.

(b) The speed of the wave is 20 m/s.

(d) The least distance between two successive crests in the wave is 2.5 cm.

Choose the correct alternatives:

1. (a, b, d)

2. (a, b, c)

3. (b, c, d)

4. (a, c, d)

(2) Hint: Use the standard equation of wave.

Given, the equation is

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

Compare the equation with the standard form,

y = asin (ωt + kx + ϕ)

Step 1: Find the direction of the wave.

1. As the equation involves a positive sign with x, hence the wave is travelling from right to left. Hence, option 1 is correct.

Step 2: Find the velocity of the wave.

2. Given,

\Rightarrow

ω = 36 \Rightarrow 2 πν = 36

ν = frequency = \frac{36}{2 π} = \frac{18}{π}

k = 0.018 \Rightarrow \frac{2 π}{λ} = 0.018

\Rightarrow \frac{2 πν}{νλ} = 0.018 [∵ 2 πν = ω and νλ = ν]

\Rightarrow \frac{36}{v} = 0.018 = \frac{18}{1000}

\Rightarrow ν = 2000 cm / s = 20 m / s

Step 3: Find the frequency of the wave.

ω = 2 πν = 36

\Rightarrow v = \frac{36}{2 π} Hz = \frac{18}{π} = 5.7 Hz

Step 4: Find the wavelength of the wave.

k = \frac{2 π}{λ} = 0.018

\Rightarrow λ = \frac{2 π}{0.018} cm

= \frac{2000 π}{18} cm = \frac{20 π}{18} m = 3.48 m

Hence, the least distance between two successive crests

= λ = 3.48 m

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