An organ pipe 40cm long is open at both ends. The speed of sound in air is 360ms-1 . The frequency of the second harmonic is ____Hz.
Correct Answer: 900
Solution and Explanation
Step 1: Formula for frequency in an open organ pipe.- Fundamental frequency f1 = $\frac{v}{2L}$, where v = 360m/s, L = 0.4m.- Frequency of second harmonic f2 = 2f1.
Step 2: Calculate the frequency.
f1 = $\frac{360}{2 \times 0.4}$ = 450Hz.
f2 = 2 × 450 = 900Hz.
Final Answer: The frequency of the second harmonic is 900Hz
Top Questions on sound
- Ribbon tweeters are often used in place of horn loaded compression tweeters because:
- What is the frequency range of flute?
- Sound waves produced at the back of a loudspeaker are________________________ degrees out of phase with respect to sound waves produced in front.
- The loudspeaker driver used for low frequency range is called_____________________.
- Crystal microphones work based on:
Questions Asked in JEE Main exam
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?
- JEE Main - 2026
- Organic Chemistry
- Let \( ABC \) be an equilateral triangle with orthocenter at the origin and the side \( BC \) lying on the line \( x+2\sqrt{2}\,y=4 \). If the coordinates of the vertex \( A \) are \( (\alpha,\beta) \), then the greatest integer less than or equal to \( |\alpha+\sqrt{2}\beta| \) is:
- JEE Main - 2026
- Coordinate Geometry
- Three charges $+2q$, $+3q$ and $-4q$ are situated at $(0,-3a)$, $(2a,0)$ and $(-2a,0)$ respectively in the $x$-$y$ plane. The resultant dipole moment about origin is ___.
- JEE Main - 2026
- Electromagnetic waves
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.- JEE Main - 2026
- Complex Numbers and Quadratic Equations
- The work functions of two metals ($M_A$ and $M_B$) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of $M_A$ : $M_B$ is in the ratio of 2.642 : 1. The work functions (in eV) of $M_A$ and $M_B$ are respectively.
- JEE Main - 2026
- Dual nature of matter