MHT CET · Physics · Oscillations
A wave is given by \(Y=3 \sin 2 \pi\left(\frac{t}{0.04}-\frac{x}{0.01}\right)\) where \(Y\) is in cm . Frequency of the wave and maximum acceleration will be \(\left(\pi^2=10\right)\)
- A \(100 \mathrm{~Hz}, 4.7 \times 10^4 \mathrm{~cm} / \mathrm{s}^2\)
- B \(50 \mathrm{~Hz}, 7.5 \times 10^3 \mathrm{~cm} / \mathrm{s}^2\)
- C \(25 \mathrm{~Hz}, 4.7 \times 10^4 \mathrm{~cm} / \mathrm{s}^2\)
- D \(25 \mathrm{~Hz}, 7.5 \times 10^4 \mathrm{~cm} / \mathrm{s}^2\)
Answer & Solution
Correct Answer
(D) \(25 \mathrm{~Hz}, 7.5 \times 10^4 \mathrm{~cm} / \mathrm{s}^2\)
Step-by-step Solution
Detailed explanation
\(y=3 \sin 2 \pi\left(\frac{t}{0.04}-\frac{x}{0.01}\right)\)
Comparing with standard wave equation,
\(\begin{array}{ll}
\mathrm{y} =\mathrm{A} \sin 2 \pi\left(\mathrm{ft} \pm \frac{\mathrm{x}}{\lambda}\right) \\
\frac{\mathrm{t}}{0.04} \\
\mathrm{f}=25 \mathrm{~Hz} \\
\omega =2 \pi \mathrm{f}=\frac{2 \pi}{0.04} \\
\mathrm{a}_{\max } =\omega^2 \mathrm{~A} \\
=\frac{4 \times \pi^2 \times 3}{(0.04)^2} \\
=7.4 \times 10^4 \mathrm{~cm} / \mathrm{s}^2 \approx 7.5 \times 10^4 \mathrm{~cm} / \mathrm{s}^2
\end{array}\)
Comparing with standard wave equation,
\(\begin{array}{ll}
\mathrm{y} =\mathrm{A} \sin 2 \pi\left(\mathrm{ft} \pm \frac{\mathrm{x}}{\lambda}\right) \\
\frac{\mathrm{t}}{0.04} \\
\mathrm{f}=25 \mathrm{~Hz} \\
\omega =2 \pi \mathrm{f}=\frac{2 \pi}{0.04} \\
\mathrm{a}_{\max } =\omega^2 \mathrm{~A} \\
=\frac{4 \times \pi^2 \times 3}{(0.04)^2} \\
=7.4 \times 10^4 \mathrm{~cm} / \mathrm{s}^2 \approx 7.5 \times 10^4 \mathrm{~cm} / \mathrm{s}^2
\end{array}\)
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