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A string of linear mass density 'μ' and length 'L' is fixed at both ends and subjected to a tension 'T'.  A small mass 'm' is attached at the midpoint.  What is the lowest frequency of transverse oscillation that can be sustained by this system?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION\n✓ **Answer**: C\n## CONCEPT FRAMEWORK\n **Primary Concept**: Standing waves on a string with added mass\n **Related Topics**: Normal modes, frequency of oscillation, linear mass density, tension\n **Prerequisites**: Understanding of standing waves, wave speed on a string, and the concept of effective mass.\n## SOLUTION PATHWAY\n**Given Information**:\n  Linear mass density = $\mu$\n  Length of string = L\n  Tension = T\n  Mass at midpoint = m\n**Method & Steps**:\n 1. Determine the effective linear mass density.\n  Reasoning: The added mass at the midpoint increases the effective mass per unit length of the string. Since the mass is concentrated at the midpoint, we can consider the mass to be effectively distributed over half of the string length (L/2) on either side of the midpoint.\n  Formula/Rule: Effective linear mass density, $\mu_{eff} = \mu + \frac{2m}{L}$\n 2. Calculate the wave speed.\n  Reasoning: The wave speed on a string is determined by the tension and the linear mass density.\n  Formula/Rule: Wave speed, $v = \sqrt{\frac{T}{\mu_{eff}}} = \sqrt{\frac{T}{\mu + \frac{2m}{L}}}$\n 3. Find the lowest frequency of oscillation (fundamental frequency).\n  Reasoning: For a string fixed at both ends, the fundamental frequency corresponds to half a wavelength fitting along the length of the string.  The relationship between frequency, wavelength, and speed is given by $v = f\lambda$.\n  Formula/Rule: Fundamental frequency, $f_1 = \frac{v}{\lambda} = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu + \frac{2m}{L}}}$\n  Result: $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu + \frac{2m}{L}}}$\n  Verification: The units are consistent (Hz). The formula shows that the frequency decreases as the added mass increases, which is intuitively correct.\n ## OPTION ANALYSIS\nA :x: Incorrectly ignores the added mass.\nB :x: Incorrectly considers the mass to be distributed over the entire length of the string.\nC ✓: Correctly considers the effective mass density.\nD :x: Incorrectly distributes the mass over only half the string length and uses the wrong denominator in the effective mass density.\n ## LEARNING AIDS\n **Common Mistakes**:\n  1.Incorrectly calculating the effective mass density:  Failing to account for the distribution of the added mass properly.\n  2.Confusing the relationship between frequency, wavelength, and wave speed.\n **Quick Tips**:\n  1.Always consider the effective mass density when an additional mass is attached to a vibrating system.\n  2.Remember that the fundamental frequency for a string fixed at both ends is given by $f_1 = \frac{v}{2L}$

VENKATESWARARAO · Answer

$\frac{1}{2L}\sqrt{\frac{T}{μ}}$

VENKATESWARARAO · Answer

$\frac{1}{2L}\sqrt{\frac{T}{μ + \frac{m}{L}}}$

VENKATESWARARAO · Answer

The correct Option is "$\frac{1}{2L}\sqrt{\frac{T}{μ + \frac{2m}{L}}}$"

VENKATESWARARAO · Answer

$\frac{1}{L}\sqrt{\frac{T}{μ + \frac{m}{2L}}}$

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