Question

A particle executing simple harmonic motion has velocities
4cms(−1)
at
6cm
and
6cm−1
at
4cm
respectively from the equilibrium position.Determine the amplitude of oscillation

Answer 1

Answer on Question #41432, Physics, Mechanics | Kinematics | Dynamics

A particle executing simple harmonic motion has velocities $4\mathrm{cms}^{-1}$ at 6cm and $6\mathrm{cms}^{-1}$ at 4cm respectively from the equilibrium position. Determine the amplitude of oscillation

Solution:

Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law.

Now since $F = -kx$ is the restoring force and from Newton's law of motion force is given as $F = ma$, where $m$ is the mass of the particle moving with acceleration $a$. Thus acceleration of the particle is

$$a = \frac{F}{m} = \frac{-kx}{m}$$

but we know that acceleration $a = dv/dt = d^2 x/dt^2$.

Thus,

$$\frac{d^2 x}{dt^2} = \frac{-kx}{m}$$

This differential equation is known as the simple harmonic equation.

The solution is

$$x = A \cos(\omega_0 t + \phi)$$

where $A$ is amplitude of oscillation, $\omega_0$ and $\phi$ are all constants.

We know that velocity of a particle is given by

$$v = \frac{dx}{dt}$$

Now differentiating the displacement of particle $x$ with respect to $t$

$$v = \frac{dx}{dt} = A \omega_0 \left( -\sin(\omega_0 t + \phi) \right)$$

From trigonometry we know that

$$\sin^2 x + \cos^2 x = 1$$

Thus,

$$A^2 \sin^2(\omega_0 t + \phi) = A^2 - A^2 \cos^2(\omega_0 t + \phi) = A^2 - x^2$$

Or

$$\sin(\omega_0 t + \phi) = \sqrt{1 - \frac{x^2}{A^2}}$$

putting this in for velocity we get,

$$v = -A \omega_0 \sqrt{1 - \frac{x^2}{A^2}}$$

so it is proportional to $\omega_0$.

Thus,

$$v^2 = A^2 \omega_0^2 \left(1 - \frac{x^2}{A^2}\right) = \omega_0^2 (A^2 - x^2)$$

We have that

$$v _ {1} = 4 \mathrm {c m s} ^ {- 1},$$ $$x _ {1} = 6 \mathrm {c m},$$ $$v _ {2} = 6 \mathrm {c m s} ^ {- 1},$$ $$x _ {2} = 4 \mathrm {c m},$$

Thus,

$$\frac {v _ {1} ^ {2}}{v _ {2} ^ {2}} = \frac {A ^ {2} - x _ {1} ^ {2}}{A ^ {2} - x _ {2} ^ {2}}$$

Substitute

$$\frac {4 ^ {2}}{6 ^ {2}} = \frac {A ^ {2} - 6 ^ {2}}{A ^ {2} - 4 ^ {2}}$$ $$\frac {1 6}{3 6} = \frac {A ^ {2} - 3 6}{A ^ {2} - 1 6}$$ $$3 6 (A ^ {2} - 3 6) = 1 6 (A ^ {2} - 1 6)$$ $$2 0 A ^ {2} = 1 0 4 0$$ $$A ^ {2} = 5 2$$ $$A = 7. 2 \mathrm {c m}$$

Answer. $A = 7.2\mathrm{cm}$

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