Question

A particle of mass m is hanging with the help of an elastic string of unstretched length a and force constant $\frac{mg}{a}$. The other end is a fixed to a peg on vertical wall. String is given an additional extension of 2a in vertical downward direction by pulling the mass and released from rest. Find the maximum height reached by it during its subsequent motion above point of release. (Neglect interaction with peg if any)

Answer 1

$\frac{9}{2}a$

Answer 2

The problem involves a particle attached to an elastic string undergoing vertical motion under gravity and elastic force. Since both forces are conservative, the total mechanical energy is conserved.

1. Determine the equilibrium position: Let the unstretched length of the string be $a$ and the force constant be $k = \frac{mg}{a}$. When the particle hangs in equilibrium, the tension in the string balances the gravitational force: $T = mg$ Also, $T = kx_{eq}$, where $x_{eq}$ is the extension from the unstretched length. So, $kx_{eq} = mg$ $\left(\frac{mg}{a}\right)x_{eq} = mg$ $x_{eq} = a$ The equilibrium length of the string is $a + x_{eq} = a + a = 2a$.

2. Determine the release position: The particle is pulled an additional $2a$ downwards from the equilibrium position. So, the total extension from the unstretched length at the release point is $x_{release} = x_{eq} + 2a = a + 2a = 3a$. The total length of the string at release is $a + 3a = 4a$. The particle is released from rest, so its initial kinetic energy is zero.

3. Set up the energy conservation equation: Let's choose the release point as the reference level for gravitational potential energy ($PE_g = 0$). Let $E_{release}$ be the total mechanical energy at the release point and $E_{max}$ be the total mechanical energy at the maximum height reached. $E = KE + PE_g + PE_{elastic}$

At the release point: $KE_{release} = 0$ (released from rest) $PE_{g, release} = 0$ (reference level) $PE_{elastic, release} = \frac{1}{2} k (x_{release})^2 = \frac{1}{2} \left(\frac{mg}{a}\right) (3a)^2 = \frac{1}{2} \left(\frac{mg}{a}\right) 9a^2 = \frac{9}{2} mga$ Total energy at release: $E_{release} = 0 + 0 + \frac{9}{2} mga = \frac{9}{2} mga$

Let $h_{max}$ be the maximum height reached by the particle above the release point. At the maximum height: $KE_{max} = 0$ (velocity is zero at max height) $PE_{g, max} = mg h_{max}$

Now, we need to determine $PE_{elastic, max}$. The string's extension depends on the particle's position. The release point is $3a$ below the unstretched length point. The unstretched length point is $a$ below the peg. So, the release point is $a + 3a = 4a$ below the peg.

If the particle moves up by $h_{max}$ from the release point, its position relative to the peg is $4a - h_{max}$ downwards. The extension of the string $x$ at this position is $(4a - h_{max}) - a = 3a - h_{max}$. Important: The string can only pull; it cannot push. If $3a - h_{max} \le 0$, the string becomes slack, and $PE_{elastic} = 0$.

4. Analyze the motion in two phases (or test for slack string):

Phase 1: String remains taut (hypothetical case) If the string remains taut, $x = 3a - h_{max} > 0$. $PE_{elastic, max} = \frac{1}{2} k (3a - h_{max})^2 = \frac{1}{2} \left(\frac{mg}{a}\right) (3a - h_{max})^2$ Applying conservation of energy: $E_{release} = E_{max}$ $\frac{9}{2} mga = mg h_{max} + \frac{1}{2} \left(\frac{mg}{a}\right) (3a - h_{max})^2$ Divide by $mg$: $\frac{9}{2} a = h_{max} + \frac{1}{2a} (3a - h_{max})^2$ Multiply by $2a$: $9a^2 = 2a h_{max} + (3a - h_{max})^2$ $9a^2 = 2a h_{max} + 9a^2 - 6a h_{max} + h_{max}^2$ $0 = h_{max}^2 - 4a h_{max}$ $h_{max}(h_{max} - 4a) = 0$ This gives $h_{max} = 0$ (the release point) or $h_{max} = 4a$. If $h_{max} = 4a$, the extension $x = 3a - 4a = -a$. This implies the string would be compressed, which is not possible. Therefore, the string must become slack before reaching this height.

Phase 2: String becomes slack The string becomes slack when its extension $x = 0$. This occurs when the particle reaches the position where its distance from the peg is equal to the unstretched length $a$. Let $h_{slack}$ be the height above the release point where the string becomes slack. The position relative to the peg is $4a - h_{slack}$. For the string to become slack, $4a - h_{slack} = a$. So, $h_{slack} = 3a$.

Let's find the velocity ($v_{slack}$) of the particle at this point using energy conservation: $E_{release} = E_{slack}$ $\frac{9}{2} mga = mg h_{slack} + \frac{1}{2} m v_{slack}^2 + PE_{elastic, slack}$ At $h_{slack} = 3a$, $PE_{elastic, slack} = 0$ (string is slack). $\frac{9}{2} mga = mg (3a) + \frac{1}{2} m v_{slack}^2$ $\frac{9}{2} mga - 3mga = \frac{1}{2} m v_{slack}^2$ $\frac{3}{2} mga = \frac{1}{2} m v_{slack}^2$ $3ga = v_{slack}^2$ $v_{slack} = \sqrt{3ga}$ (upwards)

From this point ($h_{slack} = 3a$ above the release point), the particle is in free fall (string is slack). It will continue to move upwards until its velocity becomes zero under gravity. Let $h_{additional}$ be the additional height gained during free fall. Using kinematics: $v_f^2 = v_i^2 + 2gh_{additional}$ $0^2 = (\sqrt{3ga})^2 + 2(-g)h_{additional}$ (taking upward as positive, acceleration is $-g$) $0 = 3ga - 2gh_{additional}$ $2gh_{additional} = 3ga$ $h_{additional} = \frac{3}{2} a$

The total maximum height reached above the release point is $h_{max, total} = h_{slack} + h_{additional}$. $h_{max, total} = 3a + \frac{3}{2} a = \frac{6a + 3a}{2} = \frac{9}{2} a$

5. Verification: At the maximum height $h_{max, total} = \frac{9}{2} a$: Position relative to peg: $4a - \frac{9}{2}a = -\frac{1}{2}a$. This means the particle is $\frac{1}{2}a$ above the peg. Since it's above the peg, the string is definitely slack. $KE_{max} = 0$ $PE_{g, max} = mg h_{max, total} = mg \left(\frac{9}{2}a\right) = \frac{9}{2}mga$ $PE_{elastic, max} = 0$ (string is slack) Total energy $E_{max} = \frac{9}{2}mga$. This matches $E_{release}$.