SATHEE: Oscillations - Result Question 36

Oscillations - Result Question 36

40. A particle executes linear simple harmonic motion with an amplitude of $3 c m$ . When the particle is at $2 c m$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

[2017]

(a) $\frac{\sqrt{5}}{2 π}$

(b) $\frac{4 π}{\sqrt{5}}$

(d) $\frac{\sqrt{5}}{π}$

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Answer:

Correct Answer: 40. (b)

Solution:

(b) Given, Amplitude $A = 3 c m$

When particle is at $x = 2 c m$

According to question, magnitude of velocity $=$ acceleration

$ω \sqrt{A^{2} - x^{2}} = x ω^{2}$

$\sqrt{(3)^{2} - (2)^{2}} = 2 (\frac{2 π}{T})$

$\sqrt{5} = \frac{4 π}{T} \Rightarrow T = \frac{4 π}{\sqrt{5}}$

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Oscillations - Result Question 36

40. A particle executes linear simple harmonic motion with an amplitude of 3cm. When the particle is at 2cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Answer:

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40. A particle executes linear simple harmonic motion with an amplitude of $3 c m$ . When the particle is at $2 c m$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is