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P the Maximum Speed and Acceleration of a Particle Executing Simple Harmonic Motion Are 10 Cm S−1 and 50 Cm S−2. Find the Position(S) of the Particle When the Speed is 8 Cm S−1. - Physics

Sum

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s2. Find the position(s) of the particle when the speed is 8 cm/s.

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Solution

It is given that:
Maximum speed of the particle, \[v_{Max}\]= `10 "cm"^(- 1)`

Maximum acceleration of the particle,

\[a_{Max}\]= 50 cms−2
The maximum velocity of a particle executing simple harmonic motion is given by,
\[v_{Max} = A\omega\]
where \[\text { omega is angular frequency, and }\]
is amplitude of the particle.
Substituting the value of \[v_{Max}\]in the above expression,
we get :
 = 10     \[. . . (1)\]
\[\Rightarrow \omega^2 = \frac{100}{A^2}\] 
 
aMax = ω2A = 50 cms−1

\[\Rightarrow \omega^2 = \frac{50}{A} . . . (2)\]

\[\text { From the equations (1) and (2), we get:} \]

\[\frac{100}{A^2} = \frac{50}{A}\]

\[ \Rightarrow A = 2 cm\]

\[ \therefore \omega = \sqrt{\frac{100}{A^2}} = 5 \sec^{- 1}\]

To determine the positions where the speed of the particle is 8 ms-1, we may use the following formula:
      v2 = ω2 (A2 − y2)

where y is distance of particle from the mean position, and
                 v is velocity of the particle.

On substituting the given values in the above equation, we get:
      64 = 25 (4 − y2)

\[\Rightarrow \frac{64}{25} = 4 - y^2\]

⇒ 4 − y2 = 2.56
⇒       y2 = 1.44
⇒​       y  = \[\sqrt{1 . 44}\]

⇒        y = ± 1.2 cm   (from the mean position)

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 4 | Page 252
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