The fundamental frequency of a sonometer wire increases by 6 Hz. If its tension is increased by 44%, keeping the length constant. Then find this fundamental frequency:
Since frequency \(f \propto \sqrt{T}\), increasing tension by 44% makes \(f' = f \sqrt{1.44} = 1.2f\). Thus, the change in frequency is \(0.2f = 6 \text{ Hz}\), which gives \(f = 30 \text{ Hz}\).
A string is fixed at both ends and the vibrations of string is given by the equation \(y = 10sin(2x)cos(2t)\) where \(x, y\) are in cm and \(t\) is in second. Nearby node from left end at \(x = 0\), is at a distance
Nodes occur where the spatial amplitude term \(sin(2x) = 0\), which implies \(2x = n\pi\) or \(x = \frac{n\pi}{2}\). The closest node to the left end \(x=0\) (where \(n=1\)) is at \(x = \frac{\pi}{2}\text{ cm}\).
A closed organ pipe of length \(l = 2 \text{ m}\) is vibrating in \(2^{\text{nd}}\) overtone. The frequency of vibration if speed of sound is 340 m/s is
For a closed organ pipe, the frequency of the \(n^{\text{th}}\) overtone is \(f = (2n+1)\frac{v}{4l}\). For \(2^{\text{nd}}\) overtone (\(n=2\)), \(f = 5\frac{v}{4l} = \frac{5 \times 340}{4 \times 2} = 212.5 \text{ Hz}\).
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 \(= 335\text{ m/s}\))
A well acts as a closed organ pipe. The resonant frequencies are odd harmonics: \(f_n = (2n-1)f_1\). The difference between consecutive resonant frequencies is \(300 - 180 = 120\text{ Hz}\), which corresponds to \(2f_1\). Thus, \(f_1 = 60\text{ Hz}\).
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 \(f_0 = 60\text{ Hz}\).
Assertion (A): When a pulse on string reflects from free end, the resultant pulse is formed in such a way that slope of string at free end is zero.
Reason (R): Zero resultant slope ensures that there is no force component perpendicular to string.
At a free end, there's no transverse force, so the slope \( \frac{\text{dy}}{\text{dx}} = 0 \). The resultant pulse is formed such that the free end is a displacement antinode. This implies zero slope, ensuring no transverse force. Both Assertion and Reason are true, and R correctly explains A.
Assertion (A): Node of pressure wave is formed at the open end of an organ pipe.
Reason (R): Reflected pressure wave from an open end will have phase difference of \( \pi \) w.r.t. to the incident pressure wave.
At the open end of an organ pipe, the pressure is approximately atmospheric pressure, meaning there is no excess pressure variation, hence it is a pressure node. When a pressure wave reflects from an open end (a boundary to a less dense medium), it undergoes a \( \pi \) (180 degrees) phase shift. This phase shift causes the incident and reflected waves to destructively interfere at the open end, creating a pressure node. Both A and R are true, and R explains A.
Assertion (A): Sound produced by an open organ pipe has good quality than sound produced by a closed organ pipe.
Reason (R): In OOP both even & odd harmonics are present.
The sound quality of an instrument depends on the number and intensity of overtones present. Because open organ pipes produce a full series of both even and odd harmonics, they generate a richer, higher-quality sound compared to closed pipes, which only produce odd harmonics. Therefore, both (A) and (R) are true, but (R) describes the richness of the open pipe rather than serving as the direct reason for the comparison.
Assertion (A): The fundamental frequency of an open organ pipe increases as the temperature is increased.
Reason (R): As the temperature increases, the velocity of sound increases more rapidly than length of the pipe.
The fundamental frequency of an open organ pipe is \(f = \frac{v}{2L}\). Velocity of sound \(v\) increases with temperature as \(v \propto \sqrt{T}\). While the pipe's length \(L\) also increases with temperature, the increase in \(v\) is proportionally greater than \(L\). Thus, \(f\) increases. Both A and R are true, and R correctly explains A.
Assertion (A): For a closed organ resonating pipe, the first resonance length is 60 cm. The second resonating length will be 180 cm.
Reason (R): For a particular closed pipe \(n_2 = 3n_1\).
For a closed organ pipe, the resonance lengths are in the ratio \(L_1 : L_2 : L_3 = 1 : 3 : 5\). If \(L_1 = 60 text{ cm}\), then \(L_2 = 3 times 60 \text{ cm} = 180 \text{ cm}\). So (A) is true. The resonant frequencies for a closed pipe are \(f_n = (2n-1)f_1\), thus the second resonance (third harmonic) is \(f_2 = 3f_1\). (R) is true and correctly explains (A).