Question

The equation of transverse wave on a rope is

y(x, t) = 7 sin (4.0t − 0.02x)

where y and x are measured in cm and t is expressed in second. Calculate the

maximum speed of a particle on the rope.​

Answer 1

★Solution :-

Given equation of the transverse wave  ,

$\implies\mathtt{y = 7 sin (4.0t - 0.02x) }...(1)$

The equation of a transverse wave moving in positive x - direction  ,

$\implies\mathtt{y= A sin (\omega t - kx) }..(2)$

On comparing  equation (1) and (2) with each other we get ,

• Amplitude (A) = 7 cm
• Angular velocity (w) = 4 radian
• Wave number (k  )= 0.02  rad / cm

Maximum velocity is given by the formula,

$\implies \boxed{\mathtt{v_{max } = A \omega }}$

Now let's substitute the values in the above equation,

$\implies \mathtt{ v_{max} = 7 \times 4.0 }$

$\implies \boxed{\mathtt{ v_{max} = 28 \dfrac{cm}{s} }}$

★ Derivation:-

we know that,

$\implies \mathtt{ v = \dfrac{dy}{dt}}$

Now let's substitute the value of y in the above equation ,

$\implies \mathtt{ v = \dfrac{d[A sin (\omega t - kx)]}{dt}}$

$\implies \mathtt{ v = A\times \omega \: cos (\omega t - kx)}$

The value of velocity will be maximum only when

• cos (wt + kx ) = ± 1

$\implies \boxed{\mathtt{v_{max } = A \omega }}$