Question

The tip of one prong of a tuning fork undergoes SHM of frequency 1000 Hzand amplitude 0.40 mm. For this tip, what is the magnitude of the (a) maximum acceleration, (b) maximum velocity, (c) acceleration at tip displacement 0.20 mm, and (d) velocity at tip displacement0.20 mm?

Short answer

1. Magnitude of maximum acceleration of the block is $1.6\times {10}^4\mathrm{m}/{\mathrm{s}}^2$.
2. Maximum speed of the block is $2.5\mathrm{m}/\mathrm{s}$.
3. Acceleration at tip displacement $0.20\mathrm{mm}$is $7.9\times {10}^3\mathrm{m}/{\mathrm{s}}^2$.
4. Velocity at tip displacement 0.20 mm is 2.2 m/s.

Step-by-step solution

The given data

• The frequency of the oscillation,$f=1000\mathrm{Hz}$.
• The amplitude of the oscillation, $x_m=0.40\mathrm{mm}$or$0.40\times {10}^{-3}\mathrm{m}$ .

Understanding the concept of SHM

Using the expression of SHM for maximum velocity and maximum acceleration we can find maximum velocity and acceleration. Using the expression for velocity and acceleration at a given displacement we can find the value of displacement and acceleration.

Formula:

The maximum speed of the oscillation,$V_{max}=x_m\omega$ (i)

The magnitude of maximum acceleration of the oscillation,$\left|a_{max}\right|={\omega }^2{x}_m$ (ii)

The angular frequency of the oscillation, $\omega =2\tau \tau f$ (iii)

a) Calculation of maximum acceleration of the block

The value of angular frequency using equation (iii) can be given as:

$\omega =2\times 3.14\times 1000\mathrm{Hz}$

$=6.28\times {10}^3\mathrm{rad}/\sec$

Now, the value of the magnitude of maximum acceleration using equation (ii) is given as:

$$\begin{gathered} \left|a_{max}\right|={\left(6.28\times {10}^3\mathrm{rad}/\mathrm{s}\right)}^2\times \left(0.40\times {10}^{-3}\mathrm{m}\right) \\ =15.77\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \\ =1.6\times {10}^4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value maximum acceleration is $1.6\times {10}^4\mathrm{m}/{\mathrm{s}}^2$.

b) Calculation of maximum speed

Using equation (i), the value of maximum speed of the oscillations is given as:

$$\begin{gathered} V_{max}=\left(6.28\times {10}^3\mathrm{rad}/\mathrm{s}\right)\times \left(0.40\times {10}^{-3}\mathrm{m}\right) \\ =2.512\mathrm{m}/\mathrm{s} \\ \approx 2.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the required value of speed is 2.5 m/s .

c) Calculation of the acceleration at 0.200m

Using equation (ii), the required acceleration at the position is given as:

$$\begin{gathered} \left|a_{0.20}\right|=\left(6.28\times {10}^3\right)\times 0.20\times {10}^{-3} \\ =7.886\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \\ \approx 7.9\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of acceleration is $7.9\times {10}^3\mathrm{m}/{\mathrm{s}}^2$.

d) Calculation of the speed at 0.200m

The required value of the speed at the position 0.2 m is given as:

$$\begin{gathered} V_{0.20}=\omega x_m\sqrt{1-{\left(\frac{x}{x_m}\right)}^2} \\ =\left(6.28\times {10}^3\right)\times 0.40\times {10}^{-3}\sqrt{1-{\left(\frac{0.20}{0.40}\right)}^2} \\ =2.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of speed is 2.2 m/s .