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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

A particle performs linear S.H.M. of period 4 seconds and amplitude 4 cm. Find the time taken by it to travel a distance of 1 cm from the positive extreme position. - Physics

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प्रश्न

A particle performs linear S.H.M. of period 4 seconds and amplitude 4 cm. Find the time taken by it to travel a distance of 1 cm from the positive extreme position.

बेरीज

उत्तर

Given:

T = 4 s, A = 4 cm = 0.04 m,
x = 1 cm from extreme position = 4 − 1 = 3 cm = 0.03 m

To find: Time taken (t)

Formula: x = A sin (ωt + Φ) 

Calculation:

Particle starts from a positive extreme position.

∴ Φ = `pi/2`

From formula,

x = `"A" sin((2pi"t")/"T" + Φ)` ........`(∵ ω = (2pi)/"T")`

∴ 3 = 4 sin`((2pi"t")/4 + pi/2)`

∴ cos`((2pi)/4)"t" = 3/4` ......`[∵ sin(pi/2 + θ) = cosθ]`

∴ `(pi/2)"t" = cos^-1(3/4)`

∴ t = `2/pi xx 41.1^circ xx pi/180 = 41.4/90 = 0.46 "s"`

Time taken by it to travel a distance of 1 cm from the positive extreme position is 0.46 s.  

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Amplitude (A), Period (T) and Frequency (N) of S.H.M.
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पाठ 5: Oscillations - Short Answer II

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