Question

Jee advance previous year question
________________________A stationary wave produced in a rod is expressed as,y = 0.012sin(0.15x) sin(4.8×10⁴t) what is the minimum particle speed at x = 3.5cm ?
[Note:- that y is in cm ]​

Answer 1

• Solution

At x, t :) y = 0.012sin(0.15x) sin(4.8×10⁴t)

particle velocity 8s given by dy/dx

∴V ᵖᵃʳᵗⁱᶜˡᵉ= dy/dx

= 0.012sin(0.15x) .cos (4.8×10⁴t).(4.8×10⁴)

this is manimum ,when

cos (4.8×10⁴t) = -1

So , Vᵖᵃʳᵗⁱᶜˡᵉ = (minimum at x = 3.5)

= - [0.012×4.8×10⁴× sin(0.15×3.4 ×180/3.14)°×4.8×10⁴]

=- 0.012×4.8×10⁴×sin30°

= -576×1/2 = -288cm/s=-2.88m/s Answer

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