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.