Search any question & find its solution
Question:
Answered & Verified by Expert
One end of a string is tied to the ceiling of a lift and a load is attached ai the bottom end of the string. When the lift is moving upwards with an acceleration of $2.1 \mathrm{~ms}^{-2}$, the speed of the transverse wave at the lower end of the string is $88 \mathrm{~ms}^{-1}$. If the lift moves downwards with an acceleration of $1.9 \mathrm{~ms}^{-2}$, the speed of the transverse wave at the lower end of the string is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
Options:
Solution:
1944 Upvotes
Verified Answer
The correct answer is:
$72 \mathrm{~ms}^{-1}$
When lift is moving upwards then frequency
$\mathrm{n}_{\mathrm{u}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}+\mathrm{a}}{1}}$
and when moving downwards then frequency
$\mathrm{n}_{\mathrm{d}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}-\mathrm{a}}{1}}$; wave velocity $\mathrm{V}=\mathrm{a} \omega=\mathrm{a} \times 2 \pi(\mathrm{n})$
$\therefore 88=2.1 \times 2 \pi\left(\frac{1}{2 \pi} \times \sqrt{\frac{10+2.1}{1}}\right) \Rightarrow 1=\frac{12.1 \times 2.1 \times 2.1}{88 \times 88}$
$\cong 6.9 \times 10^{-3} \mathrm{~m}$. When lift is moving downwards then wave speed.
$V=a \omega=a \times 2 \pi(n)=1.9 \times 2 \pi \times \frac{1}{2 \pi} \sqrt{\frac{10-1.9}{6.9 \times 10^{-3}}}$
$\begin{aligned} & =1.9 \times 3.42 \times 10 \\ & \therefore \mathrm{V} \cong 72 \mathrm{~m} / \mathrm{s} .\end{aligned}$
$\mathrm{n}_{\mathrm{u}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}+\mathrm{a}}{1}}$
and when moving downwards then frequency
$\mathrm{n}_{\mathrm{d}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}-\mathrm{a}}{1}}$; wave velocity $\mathrm{V}=\mathrm{a} \omega=\mathrm{a} \times 2 \pi(\mathrm{n})$
$\therefore 88=2.1 \times 2 \pi\left(\frac{1}{2 \pi} \times \sqrt{\frac{10+2.1}{1}}\right) \Rightarrow 1=\frac{12.1 \times 2.1 \times 2.1}{88 \times 88}$
$\cong 6.9 \times 10^{-3} \mathrm{~m}$. When lift is moving downwards then wave speed.
$V=a \omega=a \times 2 \pi(n)=1.9 \times 2 \pi \times \frac{1}{2 \pi} \sqrt{\frac{10-1.9}{6.9 \times 10^{-3}}}$
$\begin{aligned} & =1.9 \times 3.42 \times 10 \\ & \therefore \mathrm{V} \cong 72 \mathrm{~m} / \mathrm{s} .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.