Question

A particle moves with simple harmonic motion in a straight line . in first t sec , after starting from rest , it travels a distance 'a' and in next t sec , it travels '2a' in same direction , then

(a) amplitude motion is 3a
(b) time period of oscillation is 5π
(c) amplitude of motion is 4a
(d)Time period of oscillation is 6π ​

Answer 1

Answer:

A(1−cosωτ)=a

A(1−cos2ωτ)=3a

cosωτ=(1−

A

a

)

cos2ωτ=(1−

A

3a

)

2(1−

A

a

)

2

−1=1−

A

3a

∵cos2θ=2cos

2

θ−1

Solving the equation

A

a

=

2

1

A=2a

cosωτ=

2

1

ωτ=

3

π

T=

ω

2π

T=6τ

Explanation:

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Answer 2

A(1−cosωτ)=a

A(1−cos2ωτ)=3a

cosωτ=(1−a/A)

cos2ωτ=(1−3a/A)

2(1−a/A)^2−1=1−3a/A ∵cos2θ=2cos^2θ−1

Solving the eequation

a/A=1/2

A=2a

cosωτ=1/2

ωτ=π/3

T=2π/ω

T=6τ

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