Question 1:
easy
In a stationary wave all the particles
Question 2:
easy
The 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:
easy
Two 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:
easy
If 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:
easy
The 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:
easy
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 \(f = 30 \text{ Hz}\).