Question :- A 660Hz running fork sets up vibration in a string clamped at both ends . the wave speed for a transverse wave on this string is 220m/s and the string vibrates in three loops . (a) Find the length of the string (b) if the maximum amplitude of a particle is 0.5cm. write a suitable equation describing the motion .
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Hi !
Solution is in this given picture .
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@Raj _____✔
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Explanation:
Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s
Wave length = 220/660 = 1/3
(a)
Number of loops = 3
L = (n/2) λ
= (3/2) * (1/3)
= 50 cm
(b)
Equation of resultant stationary wave can be given by:
y = 2Acos(2πx/λ) sin(2πvL/λ)
y = 0.5 cos(6πxm⁻¹)sin(1320πts⁻¹)
Hope it helps!
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