The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by {x}({t})= Asin\ω t+ Bcos\ωtwith \\omega=\\sqrt{\frac{K}{m}}Suppose that at time {t}=0, the position of mass is {x}(0) and velocity {v}(0), then its displacement can also be represented as x(t)=C \cos (\\omegat-\ϕ), where Cand \ϕ are : [22 Jul 2021 Shift 2]

Correct Answer is : C=√v(0)2ω2+x(0)2, ϕ=tan−1(v(0) x(0)ω)

{C}=\\sqrt{{2 {v}(0)^{2}}/{\\omega^{2}}+{x}(0)^{2}}, \ϕ=\tan ^{-1}({{v}(0)}/{{x}(0) \\omega})

{C}=\\sqrt{{2 {v}(0)^{2}}/{\\omega^{2}}+{x}(0)^{2}}, \ϕ=\tan ^{-1}({{x}(0)\\omega}/{{2v}(0) })

{C}=\\sqrt{{ {v}(0)^{2}}/{\\omega^{2}}+{x}(0)^{2}}, \ϕ=\tan ^{-1}({{x}(0)\\omega}/{{v}(0) })

{C}=\\sqrt{{ {v}(0)^{2}}/{\\omega^{2}}+{x}(0)^{2}}, \ϕ=\tan ^{-1}({{v}(0) }/{{x}(0)\\omega })

JEE Main 22 July 2021 Shift 2 Solved Paper

Section: Physics

Question : 18 of 90

Marks: +1, -0

The motion of a mass on a spring, with spring constant K is as shown in figure.

The equation of motion is given by x(t)= Asin\ω t+ Bcos\ωtwith ω=√

Suppose that at time t=0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t)=C‌cos(ωt−ϕ), where Cand ϕ are :

[22 Jul 2021 Shift 2]

C=√
2v(0)2
ω2
+x(0)2, ϕ=tan−1(
v(0)
x(0)ω
)
C=√
2v(0)2
ω2
+x(0)2, ϕ=tan−1(
x(0)ω
2v(0)
)
C=√
v(0)2
ω2
+x(0)2, ϕ=tan−1(
x(0)ω
v(0)
)
C=√
v(0)2
ω2
+x(0)2, ϕ=tan−1(
v(0)
x(0)ω
)

Validate

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