SATHEE: Wave Motion 2 Question 6

Wave Motion 2 Question 6

6. Equation of travelling wave on a stretched string of linear density $5 g / m$ is $y = 0.03 \sin (450 t - 9 x)$ , where distance and time are measured in SI units. The tension in the string is

(Main 2019, 11 Jan I)

(a) $5 N$

(b) $12.5 N$

(d) $10 N$

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Answer:

Correct Answer: 6. (b)

Solution:

Given, equation can be rewritten as,

$y = 0.03 \sin 450 (t - \frac{9 x}{450})$

We know that the general equation of a travelling wave is given as,

$y = A \sin ω (t - x / v)$

Comparing Eqs. (i) and (ii), we get

$velocity, v = \frac{450}{9} = 50 m / s$

and angular velocity, $ω = 450 r a d / s$

As, the velocity of wave on stretched string with tension $(T)$ is given as $v = \sqrt{T / μ}$

where, $μ$ is linear density

$∴ T = μ v^{2} = 5 \times 10^{- 3} \times 50 \times 50 = 12.5 N$

$(∵ given, μ = 5 g / m = 5 \times 10^{- 3} k g / m)$

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Wave Motion 2 Question 6

6. Equation of travelling wave on a stretched string of linear density 5g/m is y=0.03sin⁡(450t−9x), where distance and time are measured in SI units. The tension in the string is

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6. Equation of travelling wave on a stretched string of linear density $5 g / m$ is $y = 0.03 \sin (450 t - 9 x)$ , where distance and time are measured in SI units. The tension in the string is