MHT CET · Physics · Waves and Sound
The equation of vibration of a stretched string fixed at both ends and vibrating in
\(5^{\text {th }}\) harmonic is \(Y=3 \sin (0 \cdot 4 x) \cos (200 \pi t)\) where 'x' and 'Y' are in \(\mathrm{cm}\) and \(\mathrm{t}\) in
second. Length of the string is
- A \((10 \cdot 5) \pi \mathrm{cm}\)
- B \((8 \cdot 5) \pi \mathrm{cm}\)
- C \((12 \cdot 5) \pi \mathrm{cm}\)
- D \((4 \cdot 5) \pi \mathrm{cm}\)
Answer & Solution
Correct Answer
(C) \((12 \cdot 5) \pi \mathrm{cm}\)
Step-by-step Solution
Detailed explanation
For \(5^{\text {th }}\) harmonic, the string of length \(\mathrm{L}\) vibrates in 5 loops
\(
\frac{5 \lambda}{2}=\mathrm{L}
\)
Now \(\frac{2 \pi}{\lambda}=0.4 \quad \therefore \lambda=\frac{2 \pi}{0.4}=\frac{20 \pi}{4}=5 \pi\)
\(
\therefore L=\frac{5 \times 5 \pi}{2}=\frac{25 \pi}{2}=12.5 \pi
\)
\(
\frac{5 \lambda}{2}=\mathrm{L}
\)
Now \(\frac{2 \pi}{\lambda}=0.4 \quad \therefore \lambda=\frac{2 \pi}{0.4}=\frac{20 \pi}{4}=5 \pi\)
\(
\therefore L=\frac{5 \times 5 \pi}{2}=\frac{25 \pi}{2}=12.5 \pi
\)
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