Question

A particle is moving simple harmonically between
two extreme positions x = - 8 and x = 4. Calculate
the phase difference between x = 1 and x = 4
(2) 134
wa Ola
(4) None of these​

Answer 1

Answer:

Phase difference between given two positions are

$\Delta \theta = \frac[\pi}{3}$

Explanation:

As we know that amplitude of the motion of the particle is given as

$r_1 - r_2 = 2A$

$4 - (-8) = 2A$

$A = 6 cm$

Now mean position is given as

$x_o = \frac{A_1 + A_2}{2} = \frac{-8 + 4}{2}$

$x_o = -2 cm$

now at x = 4 cm the phase of SHM is given as

$\theta = 90 degree$

now when it is at x = 1 then its phase is given as

$sin\theta = \frac{x - x_o}{A}$

$sin\theta = \frac{1 - (-2)}{6}$

$\theta = 30 degree$

Now phase difference between two positions is given as

$\Delta \theta = \frac{\pi}{2} - \frac{\pi}{6}$

$\Delta \theta = \frac[\pi}{3}$

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Topic : SHM

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Answer 2

Answer:

Phase difference between given two positions are

Explanation:

As we know that amplitude of the motion of the particle is given as

r_1 - r_2 = 2Ar

1

−r

2

=2A

4 - (-8) = 2A4−(−8)=2A

A = 6 cmA=6cm

Now mean position is given as

x_o = \frac{A_1 + A_2}{2} = \frac{-8 + 4}{2}x

o

=

2

A

1

+A

2

=

2

−8+4

x_o = -2 cmx

o

=−2cm

now at x = 4 cm the phase of SHM is given as

\theta = 90 degreeθ=90degree

now when it is at x = 1 then its phase is given as

sin\theta = \frac{x - x_o}{A}sinθ=

A

x−x

o

sin\theta = \frac{1 - (-2)}{6}sinθ=

6

1−(−2)

\theta = 30 degreeθ=30degree

Now phase difference between two positions is given as

\Delta \theta = \frac{\pi}{2} - \frac{\pi}{6}Δθ=

2

π

−

6

π