Question

19. A particle starts SHM at time t = 0. Its amplitude is A and angular frequency is a. At time t - 0 its
kinetic energy is E/4. Assuming potential energy to be zero at mean position, the displacement-time
equation of the particle can be written as (E = total mechanical energy of oscillation) :
57
(A) x = A sin(at +/6)
(B) x = Asinot +
I
6
701
(C) X = Asin ot +
(D) X = Asin wt +
6
6
(06​

Answer 1

Answer:

Its amplitude is A and angular frequency is omega. At time t=0 its kinetic energy is E,4. Assuming potential energy to be zero and the particle can be written as ...

Answer 2

Complete question:

A particle stars SHM at time t=0. Its amplitude is A and angular frequency is ω . ω. At time t=0 its kinetic energy is E 4 E4. Assuming potential energy to be zero and the particle can be written as (E=total mechanical energy of oscillation).

A. x=A cos ( ω t + π 6 ) cos(ωt+π6)

B. x = A sin ( ω t + π 3 ) sin(ωt+π3)

C. X = A sin ( ω t − 2 π 3 ) sin(ωt-2π3)

D. x = A cos ( ω t − π 6 ) cos(ωt-π6)

Answer:

All options are applicable.

Explanation:

$x = Asin(at +\frac{4\pi }{6})$

$\frac{1}{2}mv^{2} = \frac{E}{4} = \frac{1}{4} (\frac{1}{2} ma^{2} A^{2}$

⇒$V^{2} = \frac{a^{2}A^{2} }{4}$

⇒ V = ± $\frac{aA}{2}$

⇒ x= ±$\frac{\sqrt{3} A}{2}$

⇒Ф = $\frac{\pi }{3} ,\frac{2\pi }{3} ,\frac{4\pi }{3} ,\frac{5\pi }{3}$

⇒ x = $\frac{\sqrt{3} A}{2} sin(at +$Ф)

where Ф = $\frac{\pi }{3}, \frac{2\pi }{3} ,\frac{4\pi }{3} ,\frac{5\pi }{3}$

Therefore the answer is

x=A cos(ωt+π6)

x = Asin(ωt+π3)

x = A cos(ωt−π6)

X = Asin(ωt−2π3)

Hence all the given option is Applicable.

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