Question

If a body of mass 0.98 kg is made to oscillate on a spring of force constant 4.84 N/m, the angular frequency of the body is
A. 1.22 rad/s
B. 2.22 rad/s
C. 3.22 rad/s
D. 4.22 rad/s

Answer 1

Hint:The numerical problem is from the oscillations and waves section of physics. We have to use the angular frequency equation to solve the problem. The mass of the body and force constant of the spring are given in the question. We can directly substitute those values and calculate the angular frequency of the given body.

Formula Used:
Equation for angular frequency:
$\omega = \sqrt {\dfrac{k}{m}} $
Where $\omega $= angular frequency, $k$= force constant of the spring and $m$=mass of the body.

Complete step by step solution:
The equation for angular frequency:
$\omega = \sqrt {\dfrac{k}{m}} $
Where $\omega $= angular frequency, $k$= force constant of the spring and $m$=mass of the body.
The force constant of the spring, $k = 4.84N{m^{ - 1}}$
The mass of the body, $m = 0.98kg$
Then the angular frequency is given by,
$\omega = \sqrt {\dfrac{k}{m}} \\
\Rightarrow \omega = \sqrt {\dfrac{{4.84}}{{0.98}}} \\
\Rightarrow \omega = \sqrt {4.9} \\
\therefore \omega = 2.22$ rad/s

Hence, the correct option is option B.

Additional Information: The angular displacement of a wave element per unit of time is known as angular frequency. When an object travels in a circle or oscillates, it has angular frequency. The unit of angular frequency is radians per second (rad/s). The angular frequency has magnitude only (scalar) and angular velocity has both direction and magnitude (vector). The magnitude of angular velocity is known as angular frequency.

Note: Both the angular frequency and the angular velocity will be equal in a circular motion. The same symbol will be used. In contrast, the angular frequency and the angular velocity will be different in the case of simple harmonic motion. Both quantities will share the same symbol.