Question

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium position and released. At an instant 0.12 s later, its displacement has decreased to 15 cm, without the mass having passed through equilibrium. Calculate the period. (Answer: 0.67 s).

Answer 1

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$, of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$, $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

So

$x = 3.5\cos(\omega t)$

At $t = 0.12$, $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$, is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$