Vibrations of Stretched Strings
ICSE · Class 11 · Physics
Flashcards for Vibrations of Stretched Strings — ICSE Class 11 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
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Get startedState 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…
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…
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…
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…
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.
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…
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.
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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