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NEET PHYSICSMedium

A uniform rope of length LL and mass m1m_1 hangs vertically from a rigid support. A block of mass m2m_2 is attached to the free end of the rope. A transverse pulse of wavelength λ1\lambda_1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2\lambda_2. The ratio λ2λ1\frac{\lambda_2}{\lambda_1} is:

A

m1+m2m1\sqrt{\frac{m_1+m_2}{m_1}}

B

m2m1\sqrt{\frac{m_2}{m_1}}

C

m1+m2m2\sqrt{\frac{m_1+m_2}{m_2}}

D

m1m2\sqrt{\frac{m_1}{m_2}}

Step-by-Step Solution

  1. Wave Speed on a String: The speed vv of a transverse wave on a string under tension TT is given by v=Tμv = \sqrt{\frac{T}{\mu}}, where μ\mu is the linear mass density of the string .
  2. Relationship between Wavelength and Tension: Since the frequency ff of the wave remains constant as it propagates, the wavelength λ\lambda is directly proportional to the wave speed vv (because v=fλv = f\lambda). Thus, λT\lambda \propto \sqrt{T}.
  3. Tension at the Bottom (T1T_1): At the lower end of the rope, the tension is entirely due to the weight of the attached block of mass m2m_2. Therefore, T1=m2gT_1 = m_2g.
  4. Tension at the Top (T2T_2): At the upper end (at the rigid support), the tension must support the weight of both the rope and the attached block. Therefore, T2=(m1+m2)gT_2 = (m_1 + m_2)g.
  5. Calculate the Ratio: Using the proportionality λT\lambda \propto \sqrt{T}, the ratio of the wavelengths is: λ2λ1=T2T1=(m1+m2)gm2g=m1+m2m2\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2)g}{m_2g}} = \sqrt{\frac{m_1 + m_2}{m_2}}
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