Physics Oscillations

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New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

| F ? | = | I ( L ? * B ? ) |

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

? E ? d s ? = 0 ? net   = ? in   - ? out   = 0

? in   = ? out  

New answer posted

4 weeks ago

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V
Vishal Baghel

Contributor-Level 10

(A) sin (ωt) + cos (ωt) = √2 sin (ωt + π/4) ⇒ T = 2π/ω
(B) sin² (ωt) = 1/2 - (1/2)cos (2ωt) ⇒ T = 2π/ (2ω) = π/ω
(C) 3cos (π/4 - 2ωt) ⇒ T = 2π/ (2ω) = π/ω
(D) cos (ωt) + cos (2ωt) + cos (3ωt)
Time period of cos (ωt) = 2π/ω
Time period of cos (2ωt) = 2π/ (2ω)
Time period of cos (3ωt) = 2π/ (3ω)
Time period of combined function = 2π/ω

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

While the particle moves from mean position to displacement, half of its amplitude, its phase changes by π/6 rad. So,
Time taken, t = (π/6)/ω = T/12 = (2/12)s = (1/6)s
a = 6

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

A? √* (K? /m)* = A? √* (K? /m)* ⇒ A? √K? = A? √K?

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a month ago

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A
alok kumar singh

Contributor-Level 10

Given the decay equation A = A? e^ (-bt/m):
-bt/m = ln (A/A? )
Solving for b:
b = (-m/t? ) * ln (A/A? ) = (-1 / (2 * 60) * ln (6/12) = 5.775 * 10? ³ kg/s

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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a month ago

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

K = U

½ mω² (A² - x²) = ½ mω²x²

A² - x² = x²

A² = 2x²

x = ± A/√2

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

f = V (n)/ (4 (1 - 17/100) (as closed from end)
f = (n - 2) (v)/ (4 (1 - 24.5/100)
nV (100)/4 (83) = (n - 2) (v) (100)/ (4) (75.5)
75.5n = 83 (n - 2)
75.5n = 83 (M - 2)
7.5n = 166
n = 22 (approx)
f = (330) (22) (100)/ (4 (83) = 2200 Hz

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