Question

An object executes simple harmonic motion along the x-direction with angular frequency \(\omega\) and amplitude \(a\). The speed of the object is 4 cm/s and 2 cm/s when it is at distances 2 cm and 6 cm, respectively, from the equilibrium position. Which of the following is/are correct?

• A. \(\omega = \sqrt{\dfrac{3}{8}}\, \text{rad/s}\)
• B. \(\omega = \sqrt{\dfrac{5}{6}}\, \text{rad/s}\)
• C. \(a = \sqrt{\dfrac{140}{3}}\, \text{cm}\)
• D. \(a = \sqrt{\dfrac{175}{6}}\, \text{cm}\)

Hint:

In SHM, the velocity at any displacement is \(v = \omega \sqrt{a^2 - x^2}\). Using two different points allows finding both \(a\) and \(\omega\).

Answer 1

The Correct Option is A, C

Step 1: Use SHM velocity equation.
\[ v = \omega \sqrt{a^2 - x^2} \] At \(x_1 = 2\, \text{cm}, v_1 = 4\, \text{cm/s}\); At \(x_2 = 6\, \text{cm}, v_2 = 2\, \text{cm/s}\).

Step 2: Take ratio.
\[ \frac{v_1^2}{v_2^2} = \frac{a^2 - x_1^2}{a^2 - x_2^2} $\Rightarrow$ \frac{16}{4} = \frac{a^2 - 4}{a^2 - 36} \] \[ 4(a^2 - 36) = a^2 - 4 $\Rightarrow$ 3a^2 = 140 $\Rightarrow$ a = \sqrt{\frac{140}{3}}\, \text{cm} \]

Step 3: Find \(\omega\).
\[ v_1 = \omega \sqrt{a^2 - x_1^2} $\Rightarrow$ 4 = \omega \sqrt{\frac{140}{3} - 4} $\Rightarrow$ \omega = \sqrt{\frac{5}{6}}\, \text{rad/s} \]

Step 4: Conclusion.
Hence, \(a = \sqrt{\dfrac{140}{3}}\, \text{cm}\) and \(\omega = \sqrt{\dfrac{5}{6}}\, \text{rad/s}\).