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प्रश्न
Two parallel S.H.M.s represented by `"x"_1 = 5 sin(4π"t" + π//3)` cm and `"x"_2 = 3sin (4π"t" + π//4)` cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M.
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उत्तर
Given:
Two parallel S.H.Ms represented by `"x"_1 = 5 sin(4π"t" + π//3)` cm and `"x"_2 = 3 sin(4π"t" + π//4)` cm are superposed on a particle.
To find: The amplitude and epoch of the resultant SHM.
using superposition principle,
amplitude of resultant is given by,
A = `sqrt("A"_1^2 + "A"_2^2 + 2"A"_1"A"_2 "cos" phi)`
here, A₁ = 5, A₂ = 3 and Φ = `π/3 - π/4 = π/12`
now A = `sqrt(5^2 + 3^2 + 2(5)(3)cos(π/12))`
`= sqrt(25 + 9 + 30 xx 0.966)`
`= sqrt(62.98)`
= 7.936 cm
epoch of the resultant, θ = tan¯¹ `[("A"_1 "sin" Φ_1 + "A"_2 "sin" Φ_2)/("A"_1 "cos" Φ_1 + "A"_2 cos Φ_2)]`
= tan¯¹ `[(5sin(π/3) + 3sin(π/4))/(5cos(π/3) + 3cos(π/4))]`
= tan¯¹`[(5 × sqrt3/2 + 3 × 1/sqrt2)/(5 × 1/2 + 3 × 1/sqrt2)]`
= tan¯¹ `[(5 sqrt3 + 3 sqrt2)/(5 + 3sqrt2)]`
= 54° 23'
Therefore, the amplitude is 7.936 cm, and the epoch of the resultant is 54°23'.
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