Key Idea Find the parameters and put in the general wave equation. Here, A direction v=2 cm=+ ve direction =128 ms−1 and 5λ=4 Now, k=λ2π=42π×5=7.85 and v=kω=128 ms−1 ⇒ω=v×k=128×7.85=1005 As, ∴y=Asin(kx−ωt)y=2sin(7.85x−1005t)=(0.02)msin(7.85x−1005t)
A wave in a string has an amplitude of 2 ~cm. The wave travels in the +ve direction of x axis with a speed of 128 ~ms^-1 and it is noted that 5…
NEET UG 2009 — Physics Waves & Oscillations
A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x axis with a speed of 128 ms−1 and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is
Held on 30 Apr 2009 · Verified 9 Jul 2026.
Options
- A
y=(0.02)msin(7.85x+1005t)
- B
y=(0.02)msin(15.7x−2010t)
- C
y=(0.02)msin(15.7x+2010t)
- D
y=(0.02)msin(7.85x−1005t)
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