मराठी

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). Determine the tension in the string. - Physics

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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.

संख्यात्मक

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

shaalaa.com

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

shaalaa.com
Displacement Relation for a Progressive Wave
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 15: Waves - Exercises [पृष्ठ ३८८]

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एनसीईआरटी Physics [English] Class 11
पाठ 15 Waves
Exercises | Q 11.3 | पृष्ठ ३८८

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