The problem involves a particle undergoing Simple Harmonic Motion (SHM) where its velocity-squared ($v^2$) is related to its displacement ($x$) by the equation $v^2 = 50 - x^2$. The time period ($T$) is given as $T = \frac{x}{7}~\text{s}$. We need to find the displacement $x$.
SHM Equation Analysis
The standard equation for the velocity ($v$) of a particle in SHM as a function of displacement ($x$) is given by:
$v = \omega \sqrt{A^2 - x^2}$
where $\omega$ is the angular frequency and $A$ is the amplitude.
Squaring both sides, we get:
$v^2 = \omega^2 (A^2 - x^2)$
Comparing this standard form with the given equation $v^2 = 50 - x^2$, we can identify the terms:
- The coefficient of $x^2$ inside the parenthesis in the standard form is $\omega^2$. In the given equation, the coefficient of $x^2$ is $-1$. This implies that the standard form should be written as $v^2 = \omega^2 A^2 - \omega^2 x^2$.
- Comparing $v^2 = \omega^2 A^2 - \omega^2 x^2$ with $v^2 = 50 - x^2$, we get:
- $\omega^2 = 1$
- $\omega^2 A^2 = 50$
From $\omega^2 = 1$, the angular frequency is $\omega = 1$ rad/s (since frequency must be positive).
Substituting $\omega^2 = 1$ into the second equation, we get $1 \times A^2 = 50$, so $A^2 = 50$.
Time Period Calculation
The relationship between the time period ($T$) and angular frequency ($\omega$) in SHM is:
$T = \frac{2\pi}{\omega}$
Using the calculated value $\omega = 1$ rad/s:
$T = \frac{2\pi}{1} = 2\pi~\text{s}$
Solving for Displacement x
The problem states that the time period is also given by $T = \frac{x}{7}~\text{s}$.
Equating the two expressions for the time period:
$\frac{x}{7} = 2\pi$
Solving for $x$:
$x = 7 \times 2\pi$
$x = 14\pi$
To find the approximate numerical value, using $\pi \approx 3.14159$:
$x \approx 14 \times 3.14159 \approx 43.98226$
The calculated value of $x$ is approximately 44.
Similar Questions
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Important Questions from Oscillations and Waves
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