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The amplitude and the periodic time of a SHM are 5cm and 6sec respectively. At a distance of 2.5 cm away from the mean position the phase will be 1)5π/12 2)π/4 3)π/3 4)π/6?
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The amplitude and the periodic time of a SHM are 5cm and 6sec respecti...
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The amplitude and the periodic time of a SHM are 5cm and 6sec respecti...
Explanation:
Finding angular frequency:
Given, amplitude (A) = 5 cm and periodic time (T) = 6 sec
We know that,
T = 2π/ω (where ω is angular frequency)
So, ω = 2π/T = 2π/6 = π/3 rad/sec
Finding displacement:
Distance from mean position (x) = 2.5 cm
Using the equation of SHM,
x = Acos(ωt + φ)
where A is amplitude and φ is phase
Putting the given values,
2.5 = 5cos(π/3 t + φ)
cos(π/3 t + φ) = 1/2
π/3 t + φ = ±π/3 + 2nπ (where n is an integer)
π/3 t + φ = π/3 + 2nπ or π/3 t + φ = 5π/3 + 2nπ
π/3 t + φ = π/3 or π/3 t + φ = 5π/3 (taking n=0 in both cases)
φ = π/3 - π/3 t or φ = 5π/3 - π/3 t
Finding phase:
Now, we need to find the value of phase (φ) when x=2.5 cm
Putting x=2.5 cm and ω=π/3 in the above equations,
φ = π/3 - π/3 t or φ = 5π/3 - π/3 t
2.5 = 5cos(π/3 t + π/3 - φ) or 2.5 = 5cos(π/3 t + 5π/3 - φ)
2.5/5 = cos(π/3 t + π/3 - φ) or 2.5/5 = cos(π/3 t + 5π/3 - φ)
1/2 = cos(π/3 t - π/6 + φ) or 1/2 = cos(π/3 t + 7π/6 - φ)
π/3 t - π/6 + φ = π/3 + 2nπ or π/3 t - π/6 + φ = 5π/3 + 2nπ (where n is an integer)
π/3 t - π/6 + φ = 7π/6 + 2nπ or π/3 t - π/6 + φ = 11π/6 + 2nπ (where n is an integer)
π/3 t + φ = π or π/3 t + φ = 5π/3
φ = π - π/3 t or φ = 5π/3 - π/3 t
Comparing the values of φ obtained earlier and the ones obtained now, we get
φ = π - π/3 t = 5π/3 - π/3 t
2π/3 = 2π/3 t
t = 1 sec
Now, putting
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The amplitude and the periodic time of a SHM are 5cm and 6sec respectively. At a distance of 2.5 cm away from the mean position the phase will be 1)5π/12 2)π/4 3)π/3 4)π/6?
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