Question

If the displacement of a particle executing SHM be 10 cm and 12 cm when the
corresponding velocities are 16 cms
−1
and 14 cms
−1
respectively, calculate the amplitude
of motion.

Answer 1

If the displacement of a particle executing SHM is $10~\mathrm{cm}$ and $12~\mathrm{cm}$, when the corresponding velocities are $16~\mathrm{cm/s}$ and $14~\mathrm{cm/s}$ respectively, calculate the amplitude of motion.

Solution: As it is known, equations of movement and velocity of simple harmonic oscillations are:

$$x = A \cdot \sin (\omega \cdot t + \varphi_ {0}), \quad v = \frac {d x}{d t} = A \cdot \omega \cdot \cos (\omega \cdot t + \varphi_ {0}) = \omega \cdot \sqrt {A ^ {2} - [ A \sin (\omega \cdot t + \varphi_ {0}) ] ^ {2}} = \omega \cdot \sqrt {A ^ {2} - x ^ {2}};$$

Then, ratio of two different velocities is equal to: $\frac{v_2}{v_1} = \frac{\omega \cdot \sqrt{A^2 - x_2^2}}{\omega \cdot \sqrt{A^2 - x_1^2}} = \sqrt{\frac{A^2 - x_2^2}{A^2 - x_1^2}}$;

We will assume that $\frac{v_2}{v_1} = k = \frac{14}{16} = 0.875$ and determine $A$ from this equation:

$$A = \sqrt {\frac {k ^ {2} \cdot x _ {1} ^ {2} - x _ {2} ^ {2}}{k ^ {2} - 1}} = \sqrt {\frac {0 . 8 7 5 ^ {2} \cdot 0 . 1 ^ {2} - 0 . 1 2 ^ {2}}{0 . 8 7 5 ^ {2} - 1}} = 0. 1 7 0 \mathrm {m} = 1 7 \mathrm {c m};$$

Answer: 17 cm.