Question:
A string of length ℓ is fixed at both ends. It is vibrating in its 3ʳd overtone with maximum amplitude a. The amplitude at a distance ℓ/3 from one end is:
A string of length ℓ is fixed at both ends. It is vibrating in its 3ʳd overtone with maximum amplitude a. The amplitude at a distance ℓ/3 from one end is:
Show Hint
For a string fixed at both ends:
n-th overtone = (n+1)th harmonic
Always write the correct wave equation.
Updated On: Mar 19, 2026
- \(a\)
- \(0\)
- \(\dfrac{\sqrt{3}a}{2}\)
- (a)/(2)
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Third overtone corresponds to the fourth harmonic.
Step 2: Equation of stationary wave: y = asin((4π x)/(ℓ))
Step 3: At x=ℓ/3: y = asin((4π)/(3)) = a(√(3)/2)
Was this answer helpful?
0
0
Top Questions on Waves
- Two coherent loudspeakers \(L_1\) and \(L_2\) are placed at a separation of \(10\,\text{m}\) parallel to a wall at a distance of \(40\,\text{m}\) as shown in the figure. On a width \(AB\) on the wall, 10 maxima and minima are found. If the velocity of sound is \(324\,\text{m s}^{-1}\), find the frequency of sound. (Given: \( \sqrt{5}=2.23 \)).
- The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is \( \frac{5}{x} \). The value of \( x \) is ________.
- In an open organ pipe \( \nu_3 \) and \( \nu_6 \) are 3rd and 6th harmonic frequencies, respectively. If \( \nu_6 - \nu_3 = 2200\ \text{Hz} \), then the length of the pipe is _________ mm. (Take velocity of sound in air as \(330\ \text{m s}^{-1}\).)
Two loudspeakers (\(L_1\) and \(L_2\)) are placed with a separation of \(10 \, \text{m}\), as shown in the figure. Both speakers are fed with an audio input signal of the same frequency with constant volume. A voice recorder, initially at point \(A\), at equidistance to both loudspeakers, is moved by \(25 \, \text{m}\) along the line \(AB\) while monitoring the audio signal. The measured signal was found to undergo \(10\) cycles of minima and maxima during the movement. The frequency of the input signal is _____________ Hz.
(Speed of sound in air is \(324 \, \text{m/s}\) and \( \sqrt{5} = 2.23 \))
- The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be _________ cm/s.
View More Questions
Questions Asked in BITSAT exam
- The 5th term of an AP is 20 and the 12th term is 41. Find the first term.
- BITSAT - 2025
- Arithmetic Progression
- A dust particle of mass 4 × 10⁻¹² mg is suspended in air under the influence of an electric field of 50 N/C directed vertically upwards. How many electrons were removed from the neutral dust particle? (g = 10 m/s²)
- BITSAT - 2025
- Electrostatics
- Calculate the electric field at a point due to a uniformly charged spherical shell.
- BITSAT - 2025
- Electrostatics
- What is the dot product of the vectors \( \mathbf{a} = (2, 3, 1) \) and \( \mathbf{b} = (1, -1, 4) \)?
- BITSAT - 2025
- Vector Algebra
- Calculate the de Broglie wavelength of an electron moving with a given velocity.
- BITSAT - 2025
- Wave Optics
View More Questions