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Vibrations of Stretched Strings — Flashcards

ICSE · Class 11 · Physics

32 flashcards for Vibrations of Stretched Strings (ICSE Class 11 Physics) to test yourself on key terms and facts.

53 questions32 flashcards19 formulas & key relations5 concepts

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Diagram of a string fixed at both ends vibrating in its fundamental mode (first harmonic), showing one loop, two nodes at the fixed ends, and one antinode at the center. The wavelength is twice the st
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32 Flashcards·
Wave speedWave speed numericalFrequency formulaFundamental frequency numericalLaws of vibrationLaw of length numericalLaw of tension numerical
Card 1Wave speed

State the speed of a transverse wave on a stretched wire.

Answer

Formula: v = sqrt(T/m) Where: T = tension in the wire, m = mass per unit length. Meaning: greater tension increases wave speed, while greater mass per unit length decreases it. Units: v in m/s, T in n…

Card 2Wave speed numerical

A wire has tension 100 N and mass per unit length 0.01 kg/m. Find the wave speed.

Answer

Given: T = 100 N, m = 0.01 kg/m Formula: v = sqrt(T/m) Step 1: T/m = 100 / 0.01 = 10000 N·m/kg Step 2: v = sqrt(10000) = 100 m/s Answer: 100 m/s…

Card 3Frequency formula

State the frequency formula for a stretched string clamped at both ends.

Answer

Formula: n = v/(2l) = (1/(2l))sqrt(T/m) Where: n = frequency, l = vibrating length, T = tension, m = mass per unit length. Units: n in hertz, l in metre, T in newton, m in kg/m. Quick meaning: frequen…

Card 4Fundamental frequency numerical

A string of length 1.5 m vibrates with wave speed 120 m/s. Find its fundamental frequency.

Answer

Given: l = 1.5 m, v = 120 m/s Formula: n = v/(2l) Step 1: 2l = 2 x 1.5 m = 3.0 m Step 2: n = 120 / 3.0 = 40 Hz Answer: 40 Hz…

Card 5Laws of vibration

State the Law of Length for a stretched string.

Answer

Law of Length: n ∝ 1/l or nl = constant Condition: tension and mass per unit length remain constant. Meaning: a longer vibrating length gives a lower frequency. Units: n in hertz, l in metre.

Card 6Law of length numerical

A string of vibrating length 0.50 m has frequency 200 Hz. If the length is changed to 1.00 m and tension remains the same, find the new frequency.

Answer

Given: n1 = 200 Hz, l1 = 0.50 m, l2 = 1.00 m Law of Length: n1l1 = n2l2 Step 1: n2 = (n1 x l1)/l2 Step 2: n2 = (200 x 0.50)/1.00 = 100 Hz Answer: 100 Hz…

Card 7Laws of vibration

State the Law of Tension for a stretched string.

Answer

Law of Tension: n ∝ sqrt(T) Condition: length and mass per unit length remain constant. Meaning: frequency increases when tension increases. Example: tightening a string raises pitch.

Card 8Law of tension numerical

A string has frequency 150 Hz at tension 36 N. What is the frequency at tension 144 N if length and mass per unit length are unchanged?

Answer

Given: n1 = 150 Hz, T1 = 36 N, T2 = 144 N Law of Tension: n ∝ sqrt(T) So, n2/n1 = sqrt(T2/T1) Step 1: T2/T1 = 144/36 = 4 Step 2: sqrt(4) = 2 Step 3: n2 = 150 x 2 = 300 Hz Answer: 300 Hz…

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Frequently Asked Questions

What are the important topics in Vibrations of Stretched Strings for ICSE Class 11 Physics?
Key topics in Vibrations of Stretched Strings include Stretched String and Wave Speed, Laws of Vibrations of Stretched Strings, Standing Waves and Modes of Vibration, Positions of Nodes and Antinodes. Study these first, then practise questions on each for Class 11 exams.
How many flashcards are available for Vibrations of Stretched Strings?
There are 32 flashcards for Vibrations of Stretched Strings covering key definitions, facts and ideas. A few sample cards are shown on this page.
How should I revise Vibrations of Stretched Strings for Class 11 exams?
Learn the core ideas first, then work through the 53 practice questions on Vibrations of Stretched Strings. Revise definitions regularly and use flashcards for quick recall before the exam.

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