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I have the question "A mass is tethered between two springs. It is displaced by $10$ cm then released. The mass oscillates with $SHM$ with a frequency of $0.55$ Hz. Calculate the velocity when the mass is $8.0$ cm from the equilibrium position."

I used the formula

Velocity = total displacement / total time.

$$V = 0.08 / 1.82$$

And the answer I get is $V = 0.04$ m/s.

Is this correct ?

  • 0
    Are the springs identical? Are they in a horizontal, frictionless arrangement or is the mass suspended vertically between them in a vertical arrangement? – 2017-01-02
  • 0
    https://www.physicsforums.com/threads/shm-mass-between-2-prestretched-springs.213288/ – 2017-01-02

1 Answers 1

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This is simple harmonic motion.

Hence, the displacement is given by the equation:

$$s=A \cos{(\omega t)} \tag{1}$$

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

If we differentiate both sides with respect to $t$, we obtain:

$$v=-A \omega \sin{(\omega t)} \tag{2}$$

Where $v$ is the velocity.

Reminder:

$$\omega = 2\pi f=1.1\pi$$

Where $f$ is the frequency.

We evaluate a time $t$ where the displacement $s=8 \text{ cm}=0.08 \text{ m}$ using equation $(1)$.

$$0.08=0.1\cos{(\omega t)}$$ $$\frac{4}{5}=\cos{(1.1 \pi \cdot t)}$$ $$t=\frac{\arccos\left(\frac{4}{5}\right)}{1.1\pi}$$

One solution is: $$t \approx 0.186212 \text{ s}$$

We do not need to know all solutions, since for each of these solutions, the velocity is the same (Simple Harmonic Motion).

We now can substitute this value of $t$ into our expression for the velocity $v$ on equation $(2)$.

$$v=-0.1\cdot 1.1\pi \cdot \sin(1.1\pi \cdot t) \approx -0.207345 \text{ ms}^{-1}$$

Hence, the velocity when displaced by $8.0 \text{ cm}$ from the equilibrium position is:

$$|v| \approx 0.21 \text{ ms}^{-1}$$