Welcome to Lecture 105 of our comprehensive JEE Physics Master Series. In this lecture, we analyze Spring-Mass Systems in depth. We investigate the dynamics of horizontal and vertical oscillators, derive effective force constants for series and parallel spring topologies, analyze the behavior of cut and joined springs from the perspective of material elastic compliance ($k \cdot L = \text{constant}$), and introduce the two-body oscillator framework using the concept of Reduced Mass ($\mu$). The lecture concludes with 4 graded solved examples and a 10-question practice worksheet (DPP) with complete solutions.
1. Single Spring-Mass Dynamics (Horizontal & Vertical)
Consider a block of mass $m$ attached to an ideal, massless spring of stiffness constant $k$.
1. Horizontal Oscillator: On a frictionless horizontal surface, the restoring force is $F = -kx$. Newton’s second law yields:
$$m \frac{d^2 x}{dt^2} + k x = 0 \implies \omega = \sqrt{\frac{k}{m}}, \quad T = 2\pi \sqrt{\frac{m}{k}}$$
2. Vertical Oscillator: When suspended vertically in a uniform gravitational field $g$, the spring stretches by an equilibrium extension $x_0$ such that $k x_0 = m g \implies x_0 = \frac{m g}{k}$. If the mass is displaced further by $y$ from this new equilibrium position:
$$F_{\text{net}} = m g – k(x_0 + y) = m g – k x_0 – k y = -k y$$
Constant Force Invariance Principle: A constant external force (such as gravity $mg$ or an inertial pseudo-force $m a_0$) shifts the mean equilibrium position of the oscillator ($x_{\text{mean}} = F_{\text{ext}}/k$), but leaves the spring stiffness $k$, angular frequency $\omega$, and time period $T$ completely invariant.
2. Cutting and Altering Springs ($k \cdot L = \text{constant}$)
The spring constant $k$ is inversely proportional to its natural unstretched length $L$. For a homogeneous elastic wire of Young’s modulus $Y$, cross-sectional area $A$, and natural length $L$, Hooke’s law gives $F = \frac{Y A}{L} \Delta L = k \Delta L$, which implies:
$$k = \frac{Y A}{L} \implies k \cdot L = \text{constant}$$
If a uniform spring of natural length $L$ and spring constant $k$ is cut into two segments of lengths $L_1$ and $L_2$ in the ratio $L_1 : L_2 = m : n$ ($L_1 = \frac{m}{m+n}L$, $L_2 = \frac{n}{m+n}L$):
$$k_1 = \left(\frac{m+n}{m}\right) k, \quad k_2 = \left(\frac{m+n}{n}\right) k$$
A shorter spring is always stiffer than the parent spring.
3. Combinations of Springs: Series and Parallel
1. Springs in Series:
When springs are connected end-to-end, each spring experiences identical restoring tension ($F_1 = F_2 = F$), while the total extension is the sum of the individual extensions ($x_{\text{total}} = x_1 + x_2$):
$$x_{\text{total}} = \frac{F}{k_1} + \frac{F}{k_2} = \frac{F}{k_{\text{eq}}} \implies \frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} \implies k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$$
2. Springs in Parallel:
When springs are coupled side-by-side or attached to opposite sides of a mass between two rigid walls, displacement $x$ is identical for both springs, and the restoring force is additive ($F_{\text{net}} = F_1 + F_2$):
$$F_{\text{net}} = -(k_1 x + k_2 x) = -(k_1 + k_2) x \implies k_{\text{eq}} = k_1 + k_2$$
4. Two-Body Oscillations & Reduced Mass ($\mu$)
Consider two masses $m_1$ and $m_2$ resting on a frictionless surface connected by a single spring of stiffness $k$ and natural length $L_0$. If the system is released from an extended or compressed configuration without external forces, the center of mass remains at rest ($\vec{a}_{\text{cm}} = 0$).
Let coordinates of $m_1$ and $m_2$ along the line of motion be $x_1$ and $x_2$. The relative separation coordinate is $x_{\text{rel}} = x_2 – x_1$. The equation of motion for each body:
$$m_1 \frac{d^2 x_1}{dt^2} = k(x_{\text{rel}} – L_0), \quad m_2 \frac{d^2 x_2}{dt^2} = -k(x_{\text{rel}} – L_0)$$
Subtracting the acceleration equations:
$$\frac{d^2 x_{\text{rel}}}{dt^2} = \frac{d^2 x_2}{dt^2} – \frac{d^2 x_1}{dt^2} = -\left(\frac{1}{m_1} + \frac{1}{m_2}\right) k(x_{\text{rel}} – L_0)$$
Defining the reduced mass $\mu$:
$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} \implies \mu = \frac{m_1 m_2}{m_1 + m_2}$$
The relative coordinate executes SHM with equation $\mu \frac{d^2 x_{\text{rel}}}{dt^2} + k(x_{\text{rel}} – L_0) = 0$. The angular frequency and period are:
$$\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{k(m_1 + m_2)}{m_1 m_2}}, \quad T = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{m_1 m_2}{k(m_1 + m_2)}}$$
5. Solved Examples
Example 1 (Foundational – JEE Main):
A uniform spring of spring constant $k$ is cut into two pieces whose lengths are in the ratio $1 : 3$. The two pieces are then connected in parallel, and a mass $M$ is attached to the combination. Find the effective spring constant $k_{\text{eff}}$ and the time period of oscillation.
Solution:
Using $k \cdot L = \text{constant}$:
For $L_1 = \frac{1}{4}L \implies k_1 = 4k$.
For $L_2 = \frac{3}{4}L \implies k_2 = \frac{4}{3}k$.
When connected in parallel:
$$k_{\text{eff}} = k_1 + k_2 = 4k + \frac{4}{3}k = \frac{16}{3}k$$
The time period of oscillation is:
$$T = 2\pi \sqrt{\frac{M}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{M}{16k/3}} = \mathbf{\frac{\pi}{2} \sqrt{\frac{3M}{k}}}$$
Example 2 (Massive Spring Inertia Correction – JEE Main/Advanced):
An elastic spring has stiffness $k$ and non-negligible mass $m_s$. A block of mass $M$ is suspended from its lower end. Using energy conservation, determine the effective mass of the spring that participates in the oscillation, and find the modified angular frequency $\omega$.
Solution:
Let the fixed end be at $y = 0$ and the oscillating end at $y = L$ with instantaneous speed $v = \dot{y}$.
Assuming a linear distribution of velocity along the spring, an element of mass $dm = \left(\frac{m_s}{L}\right)dy$ at distance $y$ moves with speed $u(y) = \frac{y}{L} v$.
The kinetic energy of the spring is:
$$K_{\text{spring}} = \int_0^L \frac{1}{2} (dm) [u(y)]^2 = \frac{1}{2} \int_0^L \left(\frac{m_s}{L} dy\right) \left(\frac{y}{L} v\right)^2 = \frac{m_s v^2}{2 L^3} \int_0^L y^2 dy = \frac{m_s v^2}{2 L^3} \left(\frac{L^3}{3}\right) = \frac{1}{2}\left(\frac{m_s}{3}\right) v^2$$
Total kinetic energy of the system:
$$K_{\text{total}} = K_{\text{block}} + K_{\text{spring}} = \frac{1}{2} M v^2 + \frac{1}{2}\left(\frac{m_s}{3}\right) v^2 = \frac{1}{2} \left(M + \frac{m_s}{3}\right) v^2$$
The effective oscillating mass is $M_{\text{eff}} = M + \frac{m_s}{3}$.
The modified angular frequency is:
$$\omega = \mathbf{\sqrt{\frac{k}{M + \frac{m_s}{3}}}}$$
Example 3 (Pulley-Spring-Block System – JEE Advanced):
A block of mass $m$ is suspended from a massless, frictionless pulley supported by an ideal spring of stiffness $k$. The string wrapped around the pulley has one end fixed to the ceiling and the other end attached to the block. Find the time period of small vertical oscillations of the block.
Solution:
Let the downward displacement of the block from equilibrium be $x$.
Because the string wraps around the pulley, a downward displacement $x$ of the block causes the pulley to descend by $\Delta y = \frac{x}{2}$.
The spring attached to the axle of the pulley extends by $\Delta y = \frac{x}{2}$.
Potential energy stored in the spring as a function of the block’s displacement $x$:
$$U(x) = \frac{1}{2} k (\Delta y)^2 = \frac{1}{2} k \left(\frac{x}{2}\right)^2 = \frac{1}{8} k x^2 = \frac{1}{2}\left(\frac{k}{4}\right) x^2$$
Comparing this with $U(x) = \frac{1}{2} k_{\text{eff}} x^2$, the effective spring constant seen by mass $m$ is:
$$k_{\text{eff}} = \frac{k}{4}$$
The time period of vertical oscillation is:
$$T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{m}{k/4}} = \mathbf{4\pi \sqrt{\frac{m}{k}}}$$
Example 4 (Two-Body Collision and Subsequent SHM – JEE Advanced):
Two blocks of masses $m_1 = 1\text{ kg}$ and $m_2 = 2\text{ kg}$ are connected by an ideal spring of stiffness $k = 300\text{ N/m}$, resting at equilibrium on a frictionless horizontal plane. A projectile of mass $m_0 = 1\text{ kg}$ moving with speed $v_0 = 12\text{ m/s}$ collides with $m_1$ and sticks to it instantaneously. (a) Find the velocity of the center of mass of the combined system. (b) Find the time period and amplitude of the subsequent internal oscillations.
Solution:
(a) During collision with $m_1$, momentum is conserved along the line of motion:
$$m_0 v_0 = (m_0 + m_1) v_1′ \implies 1 \times 12 = (1 + 1) v_1′ \implies v_1′ = 6\text{ m/s}$$
The new total mass of block 1 is $M_1 = m_0 + m_1 = 2\text{ kg}$.
Velocity of the center of mass of the entire system ($M_1$ and $m_2$):
$$v_{\text{cm}} = \frac{M_1 v_1′ + m_2 (0)}{M_1 + m_2} = \frac{2 \times 6 + 2 \times 0}{2 + 2} = \frac{12}{4} = \mathbf{3\text{ m/s}}$$
(b) Reduced mass of the oscillating system:
$$\mu = \frac{M_1 m_2}{M_1 + m_2} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1\text{ kg}$$
Time period of oscillation:
$$T = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{1}{300}} = \mathbf{\frac{2\pi}{10\sqrt{3}}\text{ s} = \frac{\pi}{5\sqrt{3}}\text{ s}}$$
In the center-of-mass frame, the relative velocity right after collision is $v_{\text{rel}} = v_1′ – 0 = 6\text{ m/s}$.
Since the spring is at its natural length at this instant, the potential energy is zero. All internal kinetic energy transforms into maximum elastic potential energy at maximum compression/extension:
$$\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{1}{2} k A^2 \implies A = v_{\text{rel}} \sqrt{\frac{\mu}{k}} = 6 \sqrt{\frac{1}{300}} = \frac{6}{10\sqrt{3}} = \frac{\sqrt{3}}{5}\text{ m} \approx \mathbf{0.346\text{ m}}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): A spring of spring constant $k$ is cut into two equal halves. The spring constant of each half is:
(A) $k/2$
(B) $k$
(C) $2k$
(D) $4k$
Q2 (Single Choice): Two identical springs of constant $k$ are connected first in series and then in parallel to support a mass $m$. The ratio of their time periods $T_{\text{series}} / T_{\text{parallel}}$ is:
(A) $1 : 2$
(B) $2 : 1$
(C) $1 : 4$
(D) $4 : 1$
Q3 (Single Choice): A vertical spring-mass system has time period $T$ on Earth. If taken inside an elevator accelerating upward with acceleration $a = g$, the time period of vertical oscillation becomes:
(A) $T$
(B) $T/\sqrt{2}$
(C) $\sqrt{2}T$
(D) $2T$
Q4 (Single Choice): A body of mass $m$ suspended from a spring oscillates vertically with frequency $f$. If another identical spring is connected in parallel with the first spring, the new frequency of oscillation is:
(A) $f/2$
(B) $f/\sqrt{2}$
(C) $\sqrt{2}f$
(D) $2f$
Q5 (Single Choice): Two masses $m$ and $2m$ are connected by a spring of stiffness $k$ resting on a horizontal frictionless surface. The natural frequency of vibration of the system is:
(A) $\sqrt{\frac{k}{m}}$
(B) $\sqrt{\frac{k}{3m}}$
(C) $\sqrt{\frac{3k}{2m}}$
(D) $\sqrt{\frac{2k}{3m}}$
Q6 (Single Choice): A block of mass $m$ is placed between two walls and connected to each by springs of constants $k_1$ and $k_2$. If the block is displaced slightly along the line of the springs, the effective spring constant is:
(A) $k_1 + k_2$
(B) $\frac{k_1 k_2}{k_1 + k_2}$
(C) $|k_1 – k_2|$
(D) $\sqrt{k_1 k_2}$
Q7 (One or More than One Correct): Which of the following modifications will change the time period of oscillation of a horizontal spring-mass system?
(A) Increasing the amplitude of oscillation.
(B) Replacing the spring with another of double the wire radius while keeping all other dimensions identical.
(C) Performing the experiment inside an elevator moving with constant velocity.
(D) Cutting the spring to half of its original length.
Q8 (One or More than One Correct): Two blocks of masses $m_1$ and $m_2$ connected by a spring of stiffness $k$ oscillate on a smooth horizontal table without external forces. Then:
(A) The center of mass remains stationary during the oscillation.
(B) The amplitudes of oscillation of $m_1$ and $m_2$ relative to the center of mass are in the ratio $m_2 : m_1$.
(C) At any instant, the momentum of $m_1$ is equal in magnitude and opposite in direction to that of $m_2$.
(D) The frequency of oscillation depends on the total mass $(m_1 + m_2)$ alone.
Q9 (Numerical Value): A uniform spring of length $L$ and spring constant $k = 100\text{ N/m}$ is cut into two pieces in the length ratio $2 : 3$. If the smaller piece is used to oscillate a mass of $0.4\text{ kg}$, the angular frequency $\omega$ in $\text{rad/s}$ is ________.
Q10 (Numerical Value): Two masses $m_1 = 3\text{ kg}$ and $m_2 = 6\text{ kg}$ connected by a spring of spring constant $k = 1800\text{ N/m}$ oscillate on a frictionless floor. The time period of oscillation of this two-body system is $\frac{\pi}{N}\text{ seconds}$. The integer value of $N$ is ________.
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (C) | Q2: (B) | Q3: (A) | Q4: (C) | Q5: (C) | Q6: (A) | Q7: (B, D) | Q8: (A, B, C) | Q9: 25 | Q10: 15
Detailed Solutions
Solution 1:
Since $k \cdot L = \text{constant}$, halving the length ($L’ = L/2$) doubles the spring constant: $k’ = 2k$. Hence, (C) is correct.
Solution 2:
For series: $k_{\text{series}} = k/2 \implies T_{\text{series}} = 2\pi \sqrt{\frac{m}{k/2}} = 2\pi \sqrt{\frac{2m}{k}}$.
For parallel: $k_{\text{parallel}} = 2k \implies T_{\text{parallel}} = 2\pi \sqrt{\frac{m}{2k}}$.
Ratio: $\frac{T_{\text{series}}}{T_{\text{parallel}}} = \frac{\sqrt{2}}{\sqrt{1/2}} = 2 : 1$. Hence, (B) is correct.
Solution 3:
The effective gravity shifts the mean position by $x_0′ = \frac{m(g+a)}{k}$, but does not alter the restoring force derivative $\frac{dF}{dx} = -k$. The time period $T = 2\pi\sqrt{m/k}$ is invariant under constant external force fields. Hence, (A) is correct.
Solution 4:
For one spring, $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$. For two identical springs in parallel, $k_{\text{eff}} = 2k$.
$$f’ = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} = \sqrt{2} f$$
Hence, (C) is correct.
Solution 5:
For a two-body oscillator, reduced mass is $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m(2m)}{m + 2m} = \frac{2}{3}m$.
Natural frequency: $\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{k}{\frac{2}{3}m}} = \sqrt{\frac{3k}{2m}}$. Hence, (C) is correct.
Solution 6:
When the mass is displaced by $x$ to the right, spring 1 is compressed by $x$ (pushing left with force $k_1 x$) and spring 2 is stretched by $x$ (pulling left with force $k_2 x$). Both forces point in the same restoring direction:
$$F_{\text{net}} = -(k_1 + k_2) x \implies k_{\text{eff}} = k_1 + k_2$$
Hence, (A) is correct.
Solution 7:
(A) False: In simple harmonic motion, period is strictly independent of amplitude (isochronism).
(B) True: Doubling wire radius increases cross-sectional area by factor $4$, increasing $k$ and changing $T$.
(C) False: Constant velocity produces zero pseudo-force, leaving $T$ unchanged.
(D) True: Halving length doubles $k$, reducing the time period by a factor of $\sqrt{2}$.
Hence, (B, D) are correct.
Solution 8:
(A) True: Since $\Sigma F_{\text{ext}} = 0$, $a_{\text{cm}} = 0 \implies v_{\text{cm}} = \text{constant} = 0$.
(B) True: In the CM frame, $m_1 x_1 + m_2 x_2 = 0 \implies |x_1| / |x_2| = m_2 / m_1$.
(C) True: Total momentum is zero, so $\vec{p}_1 + \vec{p}_2 = 0 \implies \vec{p}_1 = -\vec{p}_2$.
(D) False: Frequency depends on the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, not total mass alone.
Hence, (A, B, C) are correct.
Solution 9:
Ratio of lengths is $2 : 3$, so the smaller piece has length $L_1 = \frac{2}{5}L$.
Spring constant: $k_1 = \frac{5}{2}k = \frac{5}{2} \times 100 = 250\text{ N/m}$.
Angular frequency with $m = 0.4\text{ kg}$:
$$\omega = \sqrt{\frac{k_1}{m}} = \sqrt{\frac{250}{0.4}} = \sqrt{625} = \mathbf{25\text{ rad/s}}$$
Answer is 25.
Solution 10:
Reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\text{ kg}$.
Time period:
$$T = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{2}{1800}} = 2\pi \sqrt{\frac{1}{900}} = 2\pi \left(\frac{1}{30}\right) = \frac{\pi}{15}\text{ s}$$
Comparing with $\frac{\pi}{N}$, we obtain $N = \mathbf{15}$.