Question

A transverse harmonic wave of amplitude 0.01 m is generated at one end of a long horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time, the displacement of a particle at a distance of 0.2 m from the fork is – 0.005 m and that of a particle at a distance of 0.1 m is + 0.005 m. The wavelength is

Options

• 0.8 m

• 0.6 m

• 0.5 m

• 0.4 m

Answer 1

0.6 m

Explanation:

y = `A sin ((2pit)/T - (2pix)/lambda)`

A = 0.01 mn = 500 Hz, T = `1/500` s

x₁ = 0.2, y₁ = – 0.005

x₂ = 0.1, y₂ = 0.005

∴ The equation is as follows: – 0.005 = `0.01 sin (2pit(500) - (2pi(0.2))/lambda)`

∴ – 0.5 = `sin(1000pit = (0.4pi)/lambda)`

∴ `1000 pit - (0.4pi)/lambda = - pi/6`  ......(i)

∴ 0.005 = `0.01 sin (2pit(500) - (2pi(0.1))/lambda)`

∴ 0.5 = `sin(1000pit - (0.2pi)/lambda)`

∴ `1000 pit - (0.2pi)/lambda = pi/6`  ......(ii)

Subtraction of equation (ii) from (i),

`100 pit - (0.4pi)/lambda - 1000 pit + (0.2pi)/lambda = (-pi)/6 - pi/6`

∴ `(-0.2pi)/lambda = (-2pi)/6`

∴ `lambda = (6 xx 0.2)/2` = 0.6 m