SATHEE: Wave Motion 5 Question 5

Wave Motion 5 Question 5

5. A string of length $0.4 m$ and mass $10^{- 2} k g$ is tightly clamped at its ends. The tension in the string is $1.6 N$ . Identical wave pulses are produced at one end at equal intervals of time $Δ t$ . The minimum value of $Δ t$ , which allows constructive interference between successive pulses, is

(1998, 2M)

(a) $0.05 s$

(b) $0.10 s$

(d) $0.40 s$

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Answer:

Correct Answer: 5. (b)

Solution:

Mass per unit length of the string,

$m = \frac{10^{- 2}}{0.4} = 2.5 \times 10^{- 2} k g / m$

$∴$ Velocity of wave in the string,

$\begin{aligned} v = \sqrt{\frac{T}{m}} = \sqrt{\frac{1.6}{2.5 \times 10^{- 2}}} \\ v = 8 m / s \end{aligned}$

For constructive interference between successive pulses

$\begin{aligned} Δ t_{min} & = \frac{2 l}{v} \\ = \frac{(2) (0.4)}{8} = 0.10 s \end{aligned}$

(After two reflections, the wave pulse is in same phase as it was produced, since in one reflection its phase changes by $π$ , and if at this moment next identical pulse is produced, then constructive interference will be obtained.)

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Wave Motion 5 Question 5

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Wave Motion 5 Question 5

5. A string of length 0.4m and mass 10−2kg is tightly clamped at its ends. The tension in the string is 1.6N. Identical wave pulses are produced at one end at equal intervals of time Δt. The minimum value of Δt, which allows constructive interference between successive pulses, is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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