Linear density of a string of is 1.3 × 10^ -4 ~kg / m and wave equation is y=0.021 (x+30 t) . Find, the tension in the string where, x in metre and t in second.

Correct answer: A. 0.12 N Mass per unit length m=1.3 × 10^ -4 ~kg / m y=0.021 (x+30 t) Comparing with y=a (ω t+(2 π x)/(λ)) a=0.021, ω=30 2 π n=30 aligned & n=(30)/(2 π) & We know that (2 π)/(λ)=1 & aligned aligned λ & =(1)/(2 π) Frequency, n & =(1)/(λ) (T)/(m) (30)/(2 π) & =(1)/(2 π) T 1.3 × 10^ -4 T & =1.3 × 10^ -4 × 900 & =0.117=0.12 ~N aligned

JIPMER 2014

Exam: JIPMER
Year: 2014
Subject: Physics
Chapter: Waves and Sound

JIPMER 2014 · Physics · Waves and Sound

Linear density of a string of is 1.3 × 10^ -4 ~kg / m and wave...

Linear density of a string of is

1.3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}

and wave equation is

y=0.021 \sin (x+30 t)

. Find, the tension in the string where,

x

in metre and

t

in second.

Options

A $0.12$ N
B $0.21$ N
C $1.2$ N
D $0.012$ N

Correct answer

A. $0.12$ N

Step-by-step solution

Mass per unit length

m=1.3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}

y=0.021 \sin (x+30 t)

Comparing with

y=a \sin \left(\omega t+\frac{2 \pi x}{\lambda}\right) a=0.021, \omega=30 2 \pi n=30

\begin{aligned} & \Rightarrow \quad n=\frac{30}{2 \pi} \\ & \text{ We know that } \frac{2 \pi}{\lambda}=1 \\ & \end{aligned}

\begin{aligned} \therefore \quad \lambda & =\frac{1}{2 \pi} \\ \text{ Frequency, } \quad n & =\frac{1}{\lambda} \sqrt{\frac{T}{m}} \\ \frac{30}{2 \pi} & =\frac{1}{2 \pi} \sqrt{\frac{T}{1.3 \times 10^{-4}}} \\ T & =1.3 \times 10^{-4} \times 900 \\ & =0.117=0.12 \mathrm{~N}\end{aligned}

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