Class 11 - Unit # 11 : Oscillations - Solved Numericals

Physics Numerical Sheet
Oscillations

1. The period of oscillation of an object in an ideal spring and mass system is 0.50 s and the amplitude is 5.0 cm. What is the speed at the equilibrium point? And the acceleration at the point of maximum extension of the spring?

Data:

• Time Period (\(T\)) = \(0.50\text{ s}\)
• Amplitude (\(x_0\)) = \(5.0\text{ cm} = 0.05\text{ m}\)
• Speed at equilibrium point (\(v_{\max}\)) = ?
• Acceleration at maximum extension (\(a_{\max}\)) = ?

Solution:

The speed of an oscillating object at any position \(x\) is given by:

$$v = \omega \sqrt{x_0^2 - x^2} \quad \text{--- (i)}$$

Since angular frequency \(\omega = \frac{2\pi}{T}\):

$$\omega = \frac{2 \times 3.14}{0.50} = 12.56\text{ rad/s}$$

At the mean/equilibrium position, \(x = 0\), rendering the speed maximum:

$$v_{\max} = \omega x_0 = 12.56 \times 0.05 = 0.628\text{ m/s}$$ $$\text{In cm/s: } v_{\max} = 0.628 \times 100 = 62.8\text{ cm/s}$$

For Acceleration, the general rule states:

$$a = -\omega^2 x \quad \text{--- (ii)}$$

At the extreme position (maximum extension), \(x = x_0\):

$$|a_{\max}| = \omega^2 x_0 = (12.56)^2 \times 0.05$$ $$|a_{\max}| = 157.75 \times 0.05 = 7.9\text{ m/s}^2$$

RESULT: The speed at the equilibrium point is 62.8 cm/s and the acceleration at maximum extension is 7.9 m/s².

2. A sewing machine needle moves with a rapid vibratory motion, like SHM, as it sews a seam. Suppose the needle moves 8.4 mm from its highest to its lowest position and it makes 24 stitches in 9.0 s. What is the maximum needle speed?

Data:

• Total path length (highest to lowest position) = \(8.4\text{ mm}\)
• Amplitude (\(x_0\)) = \(\frac{8.4}{2} = 4.2\text{ mm} = 4.2 \times 10^{-3}\text{ m}\)
• Number of vibrations/stitches = \(24\)
• Total Time = \(9.0\text{ s}\)
• Maximum speed (\(v_{\max}\)) = ?

Solution:

The peak speed of oscillation occurs at the mean position and is governed by:

$$v_{\max} = \omega x_0 = \frac{2\pi}{T} x_0 \quad \text{--- (i)}$$

The time period \(T\) for a single cycle/stitch is calculated as:

$$T = \frac{\text{Total Time}}{\text{Number of Stitches}} = \frac{9.0}{24} = 0.375\text{ s}$$

Substituting \(T\) and \(x_0\) into equation (i):

$$v_{\max} = \frac{2 \times 3.1416}{0.375} \times (4.2 \times 10^{-3})$$ $$v_{\max} = 16.755 \times 4.2 \times 10^{-3} = 0.0704\text{ m/s}$$ $$\text{In cm/s: } v_{\max} = 0.0704 \times 100 = 7.0\text{ cm/s}$$

RESULT: The maximum needle speed is 7.0 cm/s.

3. An ideal spring with a spring constant of 15 N/m is suspended vertically. A body of mass 0.60 kg is attached to the unstretched spring and released. (a) What is the extension of the spring when the speed is a maximum? (b) What is the maximum speed?

Data:

• Spring constant (\(k\)) = \(15\text{ N/m}\)
• Mass of body (\(m\)) = \(0.60\text{ kg}\)
• Acceleration due to gravity (\(g\)) = \(9.8\text{ m/s}^2\)
• Extension at maximum speed (\(x_0\)) = ?
• Maximum speed (\(v_{\max}\)) = ?

Solution:

Part (a): Speed is maximum when passing through the equilibrium point, where the downward gravitational pull equals the upward restoring force of the spring (\(F_g = F_s\)):

$$mg = k x_0 \implies x_0 = \frac{mg}{k}$$ $$x_0 = \frac{0.60 \times 9.8}{15} = \frac{5.88}{15} = 0.392\text{ m}$$

Part (b): The peak speed of oscillation is calculated using structural system parameters:

$$v_{\max} = \omega x_0 = \sqrt{\frac{k}{m}} x_0$$ $$v_{\max} = \sqrt{\frac{15}{0.60}} \times 0.392$$ $$v_{\max} = \sqrt{25} \times 0.392 = 5 \times 0.392 = 1.96\text{ m/s} \approx 2.0\text{ m/s}$$

RESULT: The static extension of the spring at equilibrium is 0.39 m and the maximum speed is 2.0 m/s.

4. A body is suspended vertically from an ideal spring of spring constant 2.5 N/m. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by \(y = (4.0\text{cm}) \sin[(0.70\text{rad/s}) t]\). What is the maximum kinetic energy of the body?

Data:

• Spring constant (\(k\)) = \(2.5\text{ N/m}\)
• Equation: \(y = (4.0\text{cm}) \sin[(0.70\text{rad/s}) t]\)
• Amplitude (\(y_0\) or \(x_0\)) = \(4.0\text{ cm} = 0.04\text{ m}\)
• Angular frequency (\(\omega\)) = \(0.70\text{ rad/s}\)
• Maximum Kinetic Energy (\(K.E_{\max}\)) = ?

Solution:

Comparing the provided expression with the generic harmonic equation \(y = x_0 \sin(\omega t)\), we confirm \(x_0 = 0.04\text{ m}\) and \(\omega = 0.70\text{ rad/s}\).

Since angular velocity relates to system characteristics via \(\omega^2 = \frac{k}{m}\), the mass element equals \(m = \frac{k}{\omega^2}\). Max kinetic energy matches total energy:

$$K.E_{\max} = \frac{1}{2} m v_{\max}^2 = \frac{1}{2} m (\omega x_0)^2 = \frac{1}{2} m \omega^2 x_0^2$$

Substituting \(k = m\omega^2\) directly simplifies the tracking energy model:

$$K.E_{\max} = \frac{1}{2} k x_0^2$$ $$K.E_{\max} = \frac{1}{2} \times 2.5 \times (0.04)^2$$ $$K.E_{\max} = 1.25 \times 0.0016 = 0.002\text{ J} = 2.0\text{ mJ}$$

RESULT: The maximum kinetic energy of the body is 2.0 mJ.

5. The period of oscillation of a simple pendulum does not depend on the mass of the bob. By contrast, the period of a mass-spring system does depend on mass. Explain this apparent contradiction.

Answer:

The period of a simple pendulum is determined solely by its length and gravitational field strength, according to the equation:

$$T = 2\pi\sqrt{\frac{l}{g}}$$

Mass does not play a role here because gravity affects all masses equally. When mass increases, the pulling force increases, but the inertia resisting the motion increases by the exact same proportion, perfectly canceling out the mass effect.

In contrast, the period of a mass-spring system is given by:

$$T = 2\pi\sqrt{\frac{m}{k}}$$

Here, the restoring force is strictly provided by the spring's elasticity (\(F = -kx\)), which stays constant regardless of the attached object. Therefore, adding mass increases the system's structural inertia without expanding the restoring force capability, causing a longer time period.

6. What is the period of a simple pendulum of a 6.0 kg mass oscillating on a 4.0 m long string?

Data:

• Mass of the bob (\(m\)) = \(6.0\text{ kg}\) (independent parameter)
• Length of string (\(l\)) = \(4.0\text{ m}\)
• Acceleration due to gravity (\(g\)) = \(9.8\text{ m/s}^2\)
• Time Period (\(T\)) = ?

Solution:

Using the pendulum period equation:

$$T = 2\pi\sqrt{\frac{l}{g}}$$ $$T = 2 \times 3.1416 \times \sqrt{\frac{4.0}{9.8}}$$ $$T = 6.2832 \times \sqrt{0.4081} = 6.2832 \times 0.6389 = 4.015\text{ s}$$

RESULT: The time period of the pendulum is 4.01 s.

7. A pendulum of length 75 cm and mass 2.5 kg swings with a mechanical energy of 0.015 J. What is its amplitude?

Data:

• Length of pendulum (\(l\)) = \(75\text{ cm} = 0.75\text{ m}\)
• Mass of bob (\(m\)) = \(2.5\text{ kg}\)
• Total mechanical energy (\(E\)) = \(0.015\text{ J}\)
• Amplitude (\(x_0\)) = ?

Solution:

Total mechanical energy equals maximum kinetic energy at the center: \(E = \frac{1}{2}mv_{\max}^2\). Since \(v_{\max} = \omega x_0\):

$$E = \frac{1}{2}m(\omega x_0)^2 = \frac{1}{2}m\omega^2 x_0^2 \quad \text{--- (i)}$$

For a pendulum, \(\omega^2 = \frac{g}{l}\). Substituting this into equation (i) yields:

$$E = \frac{1}{2}m \left(\frac{g}{l}\right) x_0^2$$ $$0.015 = \frac{1}{2} \times 2.5 \times \left(\frac{9.8}{0.75}\right) \times x_0^2$$ $$0.015 = 1.25 \times 13.066 \times x_0^2 \implies 0.015 = 16.333 \times x_0^2$$

Isolating \(x_0^2\) and taking the square root:

$$x_0^2 = \frac{0.015}{16.333} = 9.184 \times 10^{-4}$$ $$x_0 = \sqrt{9.184 \times 10^{-4}} = 0.0303\text{ m}$$ $$\text{In cm: } x_0 = 0.0303 \times 100 = 3.03\text{ cm}$$

RESULT: The amplitude of the pendulum is 3.0 cm.

8. A pendulum of length \(L_1\) has a period of \(T_1 = 0.950\text{ s}\). The length of the pendulum is adjusted to a new value \(L_2\) such that \(T_2 = 1.00\text{ s}\). What is the ratio \(L_2/L_1\)?

Data:

• Initial period (\(T_1\)) = \(0.950\text{ s}\)
• Adjusted period (\(T_2\)) = \(1.00\text{ s}\)
• Ratio of lengths (\(\frac{L_2}{L_1}\)) = ?

Solution:

The time period configuration gives us two structural cases:

$$T_1 = 2\pi\sqrt{\frac{L_1}{g}} \quad \text{--- (i)}$$ $$T_2 = 2\pi\sqrt{\frac{L_2}{g}} \quad \text{--- (ii)}$$

Dividing equation (ii) by equation (i):

$$\frac{T_2}{T_1} = \frac{2\pi\sqrt{\frac{L_2}{g}}}{2\pi\sqrt{\frac{L_1}{g}}} \implies \frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}$$

Squaring both sides to calculate the ratio:

$$\frac{L_2}{L_1} = \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{1.00}{0.950}\right)^2$$ $$\frac{L_2}{L_1} = (1.0526)^2 \approx 1.11$$

RESULT: The ratio of the lengths \(\frac{L_2}{L_1}\) is 1.11.

9. A wire is hanging from the top of a tower such that the top is not visible due to darkness. How do you calculate the height of the tower?

Answer:

The height of the tower can be determined by treating the hanging wire as a simple pendulum:

1. Attach a heavy mass securely to the lower end of the wire to turn it into a functional pendulum bob.
2. Displace the bob slightly to initiate minor small-angle horizontal vibrations.
3. Use a stopwatch to measure the total time for a fixed count of cycles (e.g., 20 oscillations) and divide by that count to calculate the precise period (\(T\)).

Now, invoke the architectural balance equation:

$$T = 2\pi\sqrt{\frac{l}{g}} \implies T^2 = 4\pi^2\frac{l}{g} \implies l = \frac{g T^2}{4\pi^2}$$

Using standard gravity (\(g = 9.8\text{ m/s}^2\)) alongside your calculated period \(T\), solve for the wire length \(l\). Adding the small remaining offset tracking distance from the bob center down to the ground provides the exact height of the tower.

10. The amplitude of oscillation of a pendulum decays by a factor of 20.0 in 120 s. By what factor has its energy decayed in that time?

Data:

• Initial Amplitude = \(x_{01}\)
• Final Amplitude after decay (\(x_{02}\)) = \(\frac{x_{01}}{20.0}\)
• Ratio of amplitudes (\(\frac{x_{01}}{x_{02}}\)) = \(20.0\)
• Energy decay ratio (\(\frac{E_1}{E_2}\)) = ?

Solution:

The total energy stored inside an oscillating system is directly proportional to the square of its path amplitude:

$$E = \frac{1}{2} m \omega^2 x_0^2 \implies E \propto x_0^2$$

Setting up a ratio comparison between the initial and final mechanical energy states:

$$\frac{E_1}{E_2} = \left(\frac{x_{01}}{x_{02}}\right)^2$$

Substituting our given amplitude decay relationship factor:

$$\frac{E_1}{E_2} = (20.0)^2 = 400$$

RESULT: The energy has decayed by a factor of 400.