One end of a string is tied to the ceiling of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the mass per unit length.

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(D) The speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the mass per unit length.At the lower end of the string, the tension $T$ is provided by the load of mass $M$. Let the mass of the string be $m$ and its length be $L$. The tension at the lower end is $T = Mg$.When the lift accelerates upwards with $a_1 = 2.1 \,ms^{-2}$, the effective acceleration is $g_{eff1} = g + a_1 = 10 + 2.1 = 12.1 \,ms^{-2}$. The tension is $T_1 = M(g + a_1)$.The wave speed is $v_1 = \sqrt{\frac{M(g+a_1)}{\mu}} = 88 \,ms^{-1}$.When the lift accelerates downwards with $a_2 = 1.9 \,ms^{-2}$, the effective acceleration is $g_{eff2} = g - a_2 = 10 - 1.9 = 8.1 \,ms^{-2}$. The tension is $T_2 = M(g - a_2)$.The wave speed is $v_2 = \sqrt{\frac{M(g-a_2)}{\mu}}$.Taking the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{g-a_2}{g+a_1}} = \sqrt{\frac{8.1}{12.1}} = \sqrt{\frac{81}{121}} = \frac{9}{11}$.Therefore, $v_2 = v_1 	imes \frac{9}{11} = 88 	imes \frac{9}{11} = 8 	imes 9 = 72 \,ms^{-1}$.

Similar Questions

$A$ uniform metal wire of density $\rho$,cross-sectional area $A$,and length $L$ is stretched with a tension $T$. The speed of a transverse wave in the wire is given by:

$A$ uniform metal wire has length $L$,mass $M$,and density $\rho$. It is under tension $T$,and $v$ is the speed of a transverse wave along the wire. The area of cross-section of the wire is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module