SATHEE: Oscillations - Result Question 39

Oscillations - Result Question 39

42. A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ , respectively. Its time period is

[2015]

(a) $2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$

(b) $2 π \sqrt{\frac{V_{1}^{2} + V_{2}^{2}}{x_{1}^{2} + x_{2}^{2}}}$

(d) $2 π \sqrt{\frac{x_{1}^{2} - x_{2}^{2}}{V_{1}^{2} - V_{2}^{2}}}$

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Answer:

Correct Answer: 42. (a)

Solution:

(a) As we know, for particle undergoing SHM,

$V = ω \sqrt{A^{2} - X^{2}}$

$V_{1}^{2} = ω^{2} (A^{2} - x_{1}^{2})$

$V_{2}^{2} = ω^{2} (A^{2} - x_{2}^{2})$

Substracting we get,

$\frac{V_{1}^{2}}{ω^{2}} + x_{1}^{2} = \frac{V_{2}^{2}}{ω^{2}} + x_{2}^{2}$

$\Rightarrow \frac{V_{1}^{2} - V_{2}^{2}}{ω^{2}} = x_{2}^{2} - x_{1}^{2}$

$\Rightarrow ω = \sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}}$

$\Rightarrow T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$

Maximum acceleration of the particle, $α = A ω^{2}$ Maximum velocity, $β = A ω$

$\Rightarrow ω = \frac{α}{β}$

$\Rightarrow T = \frac{2 π}{ω} = \frac{2 π β}{α} [∵ ω = \frac{2 π}{T}]$

Oscillations - Result Question 39

42. A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ , respectively. Its time period is

Answer:

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Oscillations - Result Question 39

42. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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42. A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ , respectively. Its time period is