Standing Wave in String and Organ Pipe - NEET Physics Chapterwise MCQs & PYQs
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NEET Standing Wave in String and Organ Pipe MCQs & PYQs
Practice NEET Standing Wave in String and Organ Pipe Questions
Question 1:
moderate
A wave in a string has an amplitude of $2text{ cm}$. The wave travels in the +ve direction of x axis with a speed of $128text{ m/sec}$ and it is noted that 5 complete waves fit in $4text{ m}$ length of the string. The equation describing the wave is
A stationary wave is represented by $y = A \sin(100t) \cos(0.01x)$, where $y$ and $A$ are in millimetres, $t$ is in seconds and $x$ is in metres. The velocity of the wave is:
(1994)
For the constituent waves, $\omega = 100$ and $k = 0.01$. Velocity $v = \frac{\omega}{k} = \frac{100}{0.01} = 10^4 \text{ m/s}$.
A wave of frequency $100 \text{ Hz}$ travels along a string towards its fixed end. When this wave travels back, after reflection, a node is formed at a distance of $10 \text{ cm}$ from the fixed end. The speed of the wave (incident and reflected) is:
(1994)
Distance from fixed end (node) to nearest node is $\frac{\lambda}{2} = 10 \text{ cm} = 0.1 \text{ m}$. Thus $\lambda = 0.2 \text{ m}$. Velocity $v = f\lambda = 100 \times 0.2 = 20 \text{ m/s}$.
The length of the string of a musical instrument is $90 \text{ cm}$ and has a fundamental frequency of $120 \text{ Hz}$. Where should it be pressed to produce fundamental frequency of $180 \text{ Hz}$?
(2020-Covid)
Fundamental frequency $f \propto \frac{1}{L}$. So $f_1 L_1 = f_2 L_2$. $120 \times 90 = 180 \times L_2 \Rightarrow L_2 = 60 \text{ cm}$.
The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \text{ cm}$, the length of the open organ pipe is:
(2018)
Fundamental of open = $v / 2L_o$. Third harmonic of closed = $3v / 4L_c$. Equating them: $v / 2L_o = 3v / 4L_c \Rightarrow L_o = 2L_c / 3 = 2(20)/3 = 13.33 \text{ cm} \approx 13.2 \text{ cm}$.
A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of $27^\circ\text{C}$ two successive resonances are produced at $20 \text{ cm}$ and $73 \text{ cm}$ of column length. If the frequency of the tuning fork is $320 \text{ Hz}$, the velocity of sound in air at $27^\circ\text{C}$ is:
The two nearest harmonics of a tube closed at one end and open at other end are $220 \text{ Hz}$ and $260 \text{ Hz}$. What is the fundamental frequency of the system?
(2017-Delhi)
For a closed pipe, frequencies are odd multiples of fundamental $f_0$. Difference between successive harmonics is $2f_0 = 260 - 220 = 40 \text{ Hz}$. Thus, $f_0 = 20 \text{ Hz}$.
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe $L$ meter long. The length of the open pipe will be:
(2016-II)
2nd overtone of open pipe = $3 \times (v / 2L_{\text{open}})$. 1st overtone of closed pipe = $3 \times (v / 4L)$. Equating them: $3v / 2L_{\text{open}} = 3v / 4L \Rightarrow L_{\text{open}} = 2L$.
An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is $50 \text{ cm}$. The next larger length of the column resonating with the same tuning fork is:
The fundamental frequency of a closed organ pipe of length $20 \text{ cm}$ is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is:
(2015)
Fundamental of closed = $v / (4 \times 20)$. 2nd overtone of open = $3v / 2L$. So, $v / 80 = 3v / 2L \Rightarrow 2L = 240 \Rightarrow L = 120 \text{ cm}$.