Q79 / 89 · KVPY 2017 (5 Nov SB/SX) · Physics
A transverse wave of frequency $500 \mathrm{~Hz}$ and speed $100 \mathrm{~m} / \mathrm{s}$ is traveling in the positive $\mathrm{x}$ direction on a long string. At time $\mathrm{t}=0 \mathrm{~s}$ the displacements at $x=0.0 \mathrm{~m}$ and at $\mathrm{x}=0.25 \mathrm{~m}$ are $0.0 \mathrm{~m}$ and $0.02 \mathrm{~m}$, respectively. The displacement at $\mathrm{x}=0.2 \mathrm{~m}$ at $\mathrm{t}=5 \times 10^{-4} \mathrm{~s}$ is
- A$-0.04 \mathrm{~m}$
- Correct answerB$-0.02 \mathrm{~m}$
- C$0.04 \mathrm{~m}$
- D$0.02 \mathrm{~m}$
Solution
$\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t})$
at $\mathrm{x}=0.025, \mathrm{y}=0.02$
$\mathrm{v}=\mathrm{v} \lambda$
$\lambda=\frac{100}{500}=\frac{1}{5} \mathrm{~m}$
$\mathrm{y}=0.02=\mathrm{A} \sin \left(\frac{2 \pi}{\lambda} \cdot \frac{1}{4}\right)$
$\mathrm{y}=0.002=\mathrm{A} \sin \left(\frac{5 \pi}{2}\right)$
$\Rightarrow \mathrm{A}=0.02 \mathrm{~m}$
$\therefore \mathrm{y}=0.02 \sin (\mathrm{kx}-\omega \mathrm{t})$
$\quad=0.02 \sin \left(10 \pi \times 0.2-1000 \pi \times 5 \times 10^{-4}\right]$
$\quad=0.02 \sin [2 \pi-0.5 \pi]$
$\quad=0.02 \sin \left(\frac{3 \pi}{2}\right)=-0.02 \mathrm{~m}$
Hence, the correct answer is option: 'B'.