Question 11:
moderate
In the given progressive wave equation y = 0.5 sin (10πt – 5x); where x, y in cm and t in second. The maximum velocity of the particle is
Question 12:
moderate
The path difference between the two waves
\[ y_{1}= a_{1} sin \left( \omega t -\frac{2\Pi x}{\lambda} \right) \]
and
\[ y_{2}= a_{2} cos \left( \omega t -\frac{2\Pi x}{\lambda} + \varphi \right) \]
is
Question 13:
moderate
The \(4^{\text{th}}\) overtone of a closed organ pipe is same as that of \(3^{\text{th}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
The frequency of the \(4^{\text{th}}\) overtone (9th harmonic) of a closed pipe is \(f_c = \frac{9v}{4L_c}\). The frequency of the \(3^{\text{rd}}\) overtone (4th harmonic) of an open pipe is \(f_o = \frac{4v}{2L_o} = \frac{2v}{L_o}\). Equating the two, \(\frac{9v}{4L_c} = \frac{2v}{L_o} ⇒ \frac{L_c}{L_o} = \frac{9}{8}\).
Question 14:
moderate
A person hums in a well and finds strong resonance at frequencies \(180\text{ Hz}\), \(300\text{ Hz}\) and \(420\text{ Hz}\). The fundamental frequency of the well is (velocity of sound = \(335\text{ m/s}\))
The resonance frequencies form an odd-harmonic progression for a closed-end pipe: \((2n-1)f_0\). The difference between consecutive harmonics is \(2f_0 = 300 - 180 = 120\text{ Hz}\) which gives \(f_0 = 60\text{ Hz}\).
Question 15:
moderate
The \(4^{\text{th}}\) overtone of a closed organ pipe is same as that of \(3^{\text{rd}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
For a closed pipe, \(f_c = \frac{9v}{4L_c}\). For an open pipe, \(f_o = \frac{4v}{2L_o}\). Since \(f_c = f_o\), we have \(\frac{9v}{4L_c} = \frac{4v}{2L_o} ⇒\frac{L_c}{L_o} = \frac{9}{8}\).
Question 16:
moderate
Velocity of sound in air is \(320\text{ m/s}\). If frequency of \(1^{\text{st}}\) overtone of a closed organ pipe is \(480\text{ Hz}\), then the length of the organ pipe is
The frequency of the \(1^{\text{st}}\) overtone (third harmonic) of a closed organ pipe is given by \(f = \frac{3v}{4L}\). Given \(f = 480\text{ Hz}\) and \(v = 320\text{ m/s}\), we have \(480 = \frac{3 \times 320}{4L} ⇒ L = \frac{960}{1920} = 0.5\text{ m} = 50\text{ cm}\).
Question 17:
moderate
The fifth overtone of a closed pipe is observed to be unison with third overtone of an open pipe. The ratio of the lengths of the pipes is
For the fifth overtone of a closed pipe, \(f_c = 11 \left(\frac{v}{4L_c}\right)\). For the third overtone of an open pipe, \(f_o = 4 \left(\frac{v}{2L_o}\right)\). Equating \(f_c = f_o\) yields \(\frac{L_c}{L_o} = \frac{11}{8}\).
Question 18:
moderate
The two nearest harmonics of an open organ pipe are 300 Hz and 450 Hz. If speed of sound in the pipe is 300 m/s, then length of the pipe is
For an open organ pipe, successive harmonics differ by the fundamental frequency: \( f_1 = 450 - 300 = 150\text{ Hz} \). Using \( f_1 = \frac{v}{2L} \), we get \( 150 = \frac{300}{2L} \implies L = 1\text{ m} = 100\text{ cm} \).
Question 19:
moderate
In a resonance tube at room temperature two successive resonance lengths of air column are \(25\text{ cm}\) and \(80\text{ cm}\). If the frequency of tuning fork is \(340\text{ Hz}\) then the speed of sound at that temperature is
The speed of sound in a resonance tube is given by \(v = 2 f (l_2 - l_1)\). Substituting \(f = 340\text{ Hz}\), \(l_1 = 0.25\text{ m}\), and \(l_2 = 0.80\text{ m}\), we get \(v = 2(340)(0.80 - 0.25) = 374\text{ m/s}\).