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NEET PHYSICSMedium

A wave in a string has an amplitude of 2 cm2 \text{ cm}. The wave travels in the +ve direction of x-axis with a speed of 128 ms1128 \text{ ms}^{-1} and it is noted that 55 complete waves fit in 4 m4 \text{ m} length of the string. The equation describing the wave is:

A

y=(0.02) msin(7.85x+1005t)y=(0.02)\text{ m} \sin(7.85x+1005t)

B

y=(0.02) msin(15.7x2010t)y=(0.02)\text{ m} \sin(15.7x-2010t)

C

y=(0.02) msin(15.7x+2010t)y=(0.02)\text{ m} \sin(15.7x+2010t)

D

y=(0.02) msin(7.85x1005t)y=(0.02)\text{ m} \sin(7.85x-1005t)

Step-by-Step Solution

  1. Determine the Amplitude (AA): The amplitude is given as 2 cm=0.02 m2 \text{ cm} = 0.02 \text{ m}.
  2. Calculate the Wavelength (λ\lambda): It is given that 55 complete waves fit in a length of 4 m4 \text{ m}. Therefore, 5λ=4 m    λ=45=0.8 m5\lambda = 4 \text{ m} \implies \lambda = \frac{4}{5} = 0.8 \text{ m}.
  3. Calculate the Wave Number (kk): The wave number is k=2πλk = \frac{2\pi}{\lambda} . k=2π0.8=2.5π2.5×3.14=7.85 rad/mk = \frac{2\pi}{0.8} = 2.5\pi \approx 2.5 \times 3.14 = 7.85 \text{ rad/m}
  4. Calculate the Angular Frequency (ω\omega): The wave speed is v=ωkv = \frac{\omega}{k}, so ω=vk\omega = v \cdot k . ω=128×7.85=1004.81005 rad/s\omega = 128 \times 7.85 = 1004.8 \approx 1005 \text{ rad/s}
  5. Determine the Wave Equation: The general equation for a wave travelling in the positive x-direction is y=Asin(kxωt)y = A\sin(kx - \omega t) or y=Asin(ωtkx)y = A\sin(\omega t - kx) . Looking at the options, we use the form y=Asin(kxωt)y = A\sin(kx - \omega t). Substituting the values, we get: y=0.02sin(7.85x1005t)y = 0.02 \sin(7.85x - 1005t)
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