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Question
A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is 220 m s−1 and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0⋅5 cm, write a suitable equation describing the motion.
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Solution
Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s
\[\text{ Wave length,} \lambda = \frac{v}{f} = \frac{220}{660} = \frac{1}{3} m\]
(a) No. of loops, n = 3
∴ \[L = \frac{n}{2}\lambda\]
\[\Rightarrow L = \frac{3}{2} \times \frac{1}{3}\]
\[ \Rightarrow L = \frac{1}{2} m = 50 \text{ cm }\]
(b) Equation of resultant stationary wave can be given by:
\[y = 2A\cos\left( \frac{2\pi x}{\lambda} \right)\sin\left( \frac{2\pi vL}{\lambda} \right)\]
\[ \Rightarrow y = 0 . 5 \cos\left( \frac{2\pi x}{\frac{1}{3}} \right)\sin\left( \frac{2\pi \times 220 \times t}{\frac{1}{3}} \right)\]
\[ \Rightarrow y = 0 . 5 cm \cos\left( 6\pi x m^{- 1} \right) \sin\left( 1320\pi t s^{- 1} \right)\]
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