A particle is subjected to two simple harmonic motions as :
$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :
SHM Components Analysis
The particle is subjected to two simple harmonic motions (SHMs) with the same angular frequency.
- SHM 1: $x_1 = \sqrt{7} \sin(5t)$ cm. Amplitude $A_1 = \sqrt{7}$ cm, Phase $\phi_1 = 0$.
- SHM 2: $x_2 = 2\sqrt{7} \sin(5t + \frac{\pi}{3})$ cm. Amplitude $A_2 = 2\sqrt{7}$ cm, Phase $\phi_2 = \frac{\pi}{3}$.
The common angular frequency is $\omega = 5$ rad/s.
The goal is to find the maximum acceleration of the particle, expressed as $x \times 10^{-2} \text{ ms}^{-2}$, and determine $x$.
Amplitude of Resultant SHM
The resultant amplitude $A$ combines individual SHMs with the same frequency using the formula:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi_2 - \phi_1)}$
Substitute the given values:
- $A_1 = \sqrt{7}$ cm
- $A_2 = 2\sqrt{7}$ cm
- $\phi_2 - \phi_1 = \frac{\pi}{3}$
- $\cos(\frac{\pi}{3}) = \frac{1}{2}$
Calculation:
$A^2 = (\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7}) \cos\left(\frac{\pi}{3}\right)$
$A^2 = 7 + (4 \times 7) + 2(2 \times 7) \left(\frac{1}{2}\right)$
$A^2 = 7 + 28 + 14$
$A^2 = 49$
$A = \sqrt{49} = 7 \text{ cm}$
Maximum Acceleration Calculation
Convert the resultant amplitude $A$ to meters:
$A = 7 \text{ cm} = 0.07 \text{ m}$
The formula for maximum acceleration ($a_{max}$) in SHM is:
$a_{max} = \omega^2 A$
Using $\omega = 5$ rad/s and $A = 0.07$ m:
$a_{max} = (5)^2 \times 0.07$
$a_{max} = 25 \times 0.07$
$a_{max} = 1.75 \text{ ms}^{-2}$
Value of x Determination
The problem states that the maximum acceleration is $x \times 10^{-2} \text{ ms}^{-2}$.
Equating the calculated value with the given format:
$1.75 = x \times 10^{-2}$
Solving for $x$:
$x = \frac{1.75}{10^{-2}} = \frac{1.75}{0.01}$
$x = 175$
Similar Questions
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Identify the correct match below :
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Important Questions from Oscillations and Waves
- Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2\text{ s}$. When fork $A$ is loaded with wax, the beat frequency is reduced to $4$ beats in $2\text{ s}$. If the original frequency of tuning fork $B$ is $380\text{ Hz}$ then original frequency of tuning fork $A$ is _________ $\text{Hz}$.
- 5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be
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The displacement of a plane progressive wave in a medium, travelling towards positive x-axis with velocity 4 m/s at t = 0 is given by \(y = 3\sin 2\pi\left(-\dfrac{x}{3}\right)\). Then the expression for the displacement at a later time t = 4 sec will be
