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प्रश्न
The transverse displacement of a string (clamped at its both ends) is given by
y(x, t) = 0.06 sin `2/3` x cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is `3.0 xx 10^(-2)` kg.
Answer the following :
Determine the tension in the string.
उत्तर १
The given equation is y(x, t) = 0.06 sin `(2π)/3 xx` cos 120 πt …(1)
Velocity of transverse waves is
`"v" = sqrt("T"/"m") or "v"^2 = "T"/"m"`
`"T" = "mv"^2, "where m" = (3xx10^(-2))/1.5 = 2xx10^(-2) "kg/m"`
`:. "T" = (180)^2 xx 2 xx 10^(-2)`
= 648 N
उत्तर २
The velocity of a transverse wave travelling in a string is given by the relation:
`v = sqrt("T"/mu)` ...(i)
Where,
Velocity of the transverse wave, v = 180 m/s
Mass of the string, m = 3.0 × 10–2 kg
Length of the string, l = 1.5 m
Mass per unit length of the string, `mu = "m"/l`
`= 3.0/1.5 xx 10^(-2)`
`= 2xx 10^(-2) "kg m"^(-1)`
Tension in the string = T
From equation (i), tension can be obtained as:
T = v2μ
= (180)2 × 2 × 10–2
= 648 N
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