Question 1:
easyIn a stationary wave all the particles
Practice NEET Standing Wave in String and Organ Pipe Questions
Question 2:
easyThe particle displacement (in cm) in a stationary wave is given by y(x, t) = 2 sin (0.1 πx) cos (100 πt). The distance between a node and the next antinode is
Question 3:
easyTwo tuning forks A and B when sounded together produce \(4\text{ beats/s}\). When B is loaded with wax, the beat frequency remains same. If frequency of A is \(212\text{ Hz}\). The frequency of B before loading is:
The frequency of B must be either \(212 + 4 = 216\text{ Hz}\) or \(212 - 4 = 208\text{ Hz}\). Loading B with wax decreases its frequency. For the beat frequency to remain \(4\text{ Hz}\), its frequency must drop from \(216\text{ Hz}\) to \(208\text{ Hz}\). Thus, the initial frequency of B is \(216\text{ Hz}\).
Question 4:
easyIf a string fixed at both ends vibrates in three loops, the wavelength is \(30\text{ cm}\). The length of string is:
For a string fixed at both ends vibrating in \(n\) loops, the length of the string is given by \(L = \frac{n\lambda}{2}\). Here, \(n = 3\) and \(\lambda = 30\text{ cm}\), so \(L = 3 \times \frac{30}{2} = 45\text{ cm}\).
Question 5:
easyThe frequency of the first overtone of a closed pipe of length \(L_1\), is equal to that of the first overtone of an open pipe of length \(L_2\). The ratio of their lengths \((L_1 : L_2)\) is:
The first overtone of a closed pipe of length \(L_1\) has frequency \(f_{c,1} = \frac{3v}{4L_1}\) and that of an open pipe of length \(L_2\) is \(f_{o,1} = \frac{v}{L_2}\). Equating the two frequencies gives \(\frac{3v}{4L_1} = \frac{v}{L_2}\), which simplifies to \(\frac{L_1}{L_2} = \frac{3}{4}\).
Question 6:
easyThe 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 \(f = 30 \text{ Hz}\).
Question 7:
easyA 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 \(x = \frac{\pi}{2}\text{ cm}\).
Question 8:
easyA 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 (\(n=2\)), \(f = 5\frac{v}{4l} = \frac{5 \times 340}{4 \times 2} = 212.5 \text{ Hz}\).
Question 9:
easyA 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, \(f_1 = 60\text{ Hz}\).
Question 10:
easyAssertion (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.