Question 159: Oscillations

Question

A particle of mass $$0.50 \mathrm{~kg}$$ executes simple harmonic motion under force $$F=-50(\mathrm{Nm}^{-1}) x$$. The time period of oscillation is $$\frac{x}{35} s$$. The value of $$x$$ is _________.
(Given $$\pi=\frac{22}{7}$$)

Accepted Answer

To find the value of $$x$$ that represents the time period of oscillation in this simple harmonic motion (SHM) scenario, we first recall the general formula for the time period ($T$) of a mass-spring system undergoing SHM, which is given by:
$$T = 2\pi \sqrt{\frac{m}{k}}$$
Here,
$m$ is the mass of the particle, which is $$0.50 \, \mathrm{kg}$$ in this case,
$k$ is the force constant of the spring or the spring constant, which is given as $$50 \, \mathrm{Nm^{-1}}$$,
and $T$ represents the time period of oscillation.
Given in the problem, $$T = \frac{x}{35} \, \mathrm{s}$$ and we are provided with the approximation $$\pi = \frac{22}{7}$$.
Substituting the given values into the formula for $$T$$:
$$\frac{x}{35} = 2 	imes \frac{22}{7} 	imes \sqrt{\frac{0.50}{50}}$$
To simplify this, we first calculate the square root:
$$\sqrt{\frac{0.50}{50}} = \sqrt{\frac{1}{100}} = \frac{1}{10}$$
Substituting back, we get:
$$\frac{x}{35} = 2 	imes \frac{22}{7} 	imes \frac{1}{10}$$
Multiplying the terms on the right side:
$$\frac{x}{35} = \frac{44}{70}$$
$$\frac{x}{35} = \frac{22}{35}$$
Multiplying both sides by $$35$$ to solve for $$x$$:
$$x = 22$$
Therefore, the value of $$x$$ that represents the time period of oscillation is $$22$$ seconds.

Oscillations

A particle of mass $0.50 \mathrm{~kg} executes simple harmonic motion under force F=-50(\mathrm{Nm}^{-1}) x. The time period of oscillation is \frac{x}{35} s. The value of x is _________. (Given \pi=\frac{22}{7}$)

Integer / Numeric Answer Type