Question
One end of a long and thin rope, stretched horizontally with a tension T = 8 N, along x axis, is supporting a weight after passing over a pulley fixed on a vertical pole (see figure). At the other end, a simple harmonic oscillator (a clamped iron rod along the axis of a solenoid fed with AC voltage and oscillating between north and south poles) at x = 0, generates a transverse wave of frequency 100 Hz and an amplitude of 2 cm, in the rope. The wave propagates along the rope. The mass per unit length of the rope is 20 g/m. Ignoring the effect of gravity (on the rope), the correct options are :-
Options
Solution
[Key Concept: Wave speed on a string is √(T/μ); frequency is set by the source, so λ = v/f.]
Velocity of wave in rope $= \sqrt{T/\mu} = \sqrt{\frac{8}{20 \times 10^{- 3}}} = 20\text{ }m/s$
a) Wavelength $= \frac{v}{f} = \frac{20}{100}\text{ }m = 20\text{ }cm$
b) Max. transverse acceleration $= \omega^{2}\text{ }A$
$$= (2\pi \times 100)^{2} \times \frac{2}{100} = 800\pi^{2}\text{ }m/s^{2}$$
c) Equation of wave $= y = Asin(\omega t - kx + \phi) = 0.02sin\left( 200\pi - \frac{2\pi}{0.2}x + \phi \right)$
$${\begin{matrix} & \text{~at~}t = 0,x = 0y = - A \\ & \ - A = Asin\phi \\ & \phi = - 90^{\circ} \end{matrix} } {so\ equation\ is\ :\ y = - 0.02cos(200\pi - 10\pi x)}$$
d) if $f = 200\text{ }Hz\lambda = \frac{v}{f} = \frac{20}{200} = 10\text{ }cm$
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