Question

during which part of the oscillation the two are along the same direction?​

Answer 1

Answer:

Two particles execute simple harmonic motion.

Let ϕ be the phase difference between the two simple harmonic motions.

The two simple harmonic motions are represented by y

1

=asinωt

y

2

=asin(ωt+ϕ)

Let the two particles meet one another, while moving in opposite direction at t=t

1

∴ At t=t

1

,y

1

=y

2

=

2

a

∴

2

a

=asinωt

1

sinωt

1

=

2

1

.....(i)

2

a

=asin(ωt

1

+ϕ)⇒sin(ωt

1

+ϕ)=

2

1

....(ii)

2

1

=sinωt

1

cosϕ+sinϕcosωt

1

=

2

1

=

2

1

cosϕ+sinϕ

1−sin

2

ωt

1

[Using (i)]

2

1

=

2

1

cosϕ+sinϕ

1−(

2

1

)

2

2

1

=

2

1

cosϕ+

2

3

sinϕ⇒1=cosϕ+

3

sinϕ

1−cosϕ=

3

sinϕ....(iii)

Square equation (iii), on both sides

1+cos

2

ϕ−2cosϕ=3sin

2

ϕ

1+cos

2

ϕ−2cosϕ=3(1−cos

2

ϕ)

4cos

2

ϕ−2cosϕ−2=0

(2cosϕ+1)(cosϕ−1)−0....(iv)

Equation (iv) is satisfied if either

2cosϕ+1=0 i.e. cosϕ=−

2

1

or ϕ=120

∘

=

3

2π

or cosϕ−1=0

cosϕ=1 or ϕ=0

ϕ=0 is not the solution to the problem.

Therefore ϕ=120

∘

.

Explanation:

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Answer 2

x0[cos(2.1t)cos(50t)]=2x0[cos52.1t+cos47.9t]

We can easily see that two constituent SHM's are y1=2x0cos52.1t and y2=2x0cos47.9t, So we have

ω1=52.1s−1

ω2=47.9s−1

and frequency of beats is given by f=2πω1−ω2 or T=ω1−ω22π

⇒T=4.22×π=1.49s≅1.5s