Question 1:
easyInstantaneous profile of a rope carrying a progressive wave moving from left to right is shown. Find the correct option
Practice NEET Progressive Wave in a String Questions
Question 2:
easyA particle moves according to equation, \(x = a \cos\frac{\pi t}{2}\). The distance covered by it in the time interval between \(t = 0\) to \(t = 3\text{ s}\) is
The time period is \(T = \frac{2\pi}{\pi/2} = 4\text{ s}\). The interval \(t = 3\text{ s}\) corresponds to \(\frac{3T}{4}\). In each quarter cycle, the distance covered is \(a\), so total distance is \(3a\).
Question 3:
moderateA transverse wave is represented by $y = A\sin(\omega t – kx)$. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?
(2010 Pre)
The maximum particle velocity is $v_{max} = A\omega$. The wave velocity is $v = \omega/k$. Equating the two gives $\omega/k = A\omega$, so $1/k = A$. Since $k = 2\pi/\lambda$, we get $\lambda/(2\pi) = A$, which yields $\lambda = 2\pi A$.
Question 4:
difficultA uniform rope of length $L$ and mass $m_1$ hangs vertically from a rigid support. A block of mass $m_2$ is attached to the free end of the rope. A transverse pulse of wavelength $\lambda_1$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is $\lambda_2$. The ratio $\lambda_2/\lambda_1$ is:
(2016 – I)
$v = f\lambda \Rightarrow \lambda \propto v \propto \sqrt{T}$. At the bottom, tension $T_1 = m_2 g$. At the top, $T_2 = (m_1+m_2)g$. Thus $\lambda_2/\lambda_1 = \sqrt{T_2/T_1} = \sqrt{\frac{m_1+m_2}{m_2}}$.
Question 5:
moderateIf the initial tension on a stretched string is doubled, then the ratio of the initial and final speed of a transverse wave along the string is :
(2022)
Speed of transverse wave $v = \sqrt{\frac{T}{\mu}}$. If $T$ is doubled, $v_{\text{final}} = \sqrt{2} v_{\text{initial}}$. Ratio $v_{\text{initial}} : v_{\text{final}} = 1 : \sqrt{2}$.
Question 6:
moderateA $5.5 \text{ metre}$ length of string has a mass of $0.035 \text{ kg}$. If the tension in the string in $77 \text{ N}$, the speed of a wave on the string is:
(1989)
Linear mass density $\mu = m/L = 0.035 / 5.5 = 0.00636 \text{ kg/m}$. Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{77}{0.035/5.5}} = \sqrt{12100} = 110 \text{ ms}^{-1}$.