More Problems

Problem 1:

A triple pendulum consists of masses am, m and m attached to a single light string at distances a, 2a, and 3a respectively from its points of suspension. Consider only motion in a plane.
(a) Determine the value of a such that one of the normal frequencies of the system will equal the frequency of a simple pendulum of length a/2 and mass m. You may assume the displacements of the masses from equilibrium are small.
(b) Find the mode corresponding to this frequency and sketch it.

Solution:

	Concepts, principles, relations that apply to the problem: Small oscillations, coupled oscillations, normal modes Solutions of the form q_j = Re(A_jeⁱ^w^t) can be found. We can find the w² from det(k_ij- w²T_ij) = 0.
	Why do they apply? We are asked to find the normal mode frequencies for small displacements.
	How do they apply? Let f₁, f₂, and f₃ be the generalized coordinates. Then x₁ = a sinf₁, x₂ = a(sinf₁ + sinf₂), x₃ = a(sinf₁ + sinf₂ + sinf₃), y₁ = a cosf₁, y₂ = a(cosf₁ + cosf₂), x₃ = a(cosf₁ + cosf₂ + cosf₃). For small displacements we have sinf = f, cosf = 1 - f²/2. The kinetic energy of the system is . Let us only keep terms up to second order in f and df/dt. Then . . The potential energy of the system is U = -amgy₁ - mgy₂- mgy₃.^.We have: . T₁₁ = ma²(a + 2), T₂₂ = 2ma², T₃₃ = ma², T₁₂ = T₂₁ = 2ma², T₁₃ = T₃₁ = T₂₃ = T₃₂ = ma². k₁₁ = mga(a + 2), k₂₂ = 2mga, k₃₃ = mga, k_ij(i ¹ j) = 0. . .
	Details of the calculations: For a simple pendulum with length a/2 we have w = (2g/a)^1/2. We are looking for a mode with w² = 2g/a, or l = 2. . The mode corresponding to this frequency is found from . A₃ = -2A₁, A₂ = 0.

Problem 2:

A uniform thin rod of length 3l/2 and mass m is supported at one end A by a weightless string of length l under the gravitational force near the earth surface.

(a) Calculate the normal modes and normal frequencies of small oscillations for motion in a vertical plane.
(b) Now the end point A is slowly displaces by a small amount d, (without gaining any kinetic energy.) The system is then released from rest and allowed to move freely. What is the subsequent motion?

Solution:

	Concepts, principles, relations that apply to the problem: Small oscillations: Solutions of the form q_j = Re(A_jeⁱ^w^t) can be found. For a particular frequency w_a we solve to find the A_j_a. The most general solution for each coordinate q_j is a sum of simple harmonic oscillations in all of the frequencies w_a. .
	Why do they apply? We are asked to find the normal modes for small displacements.
	How do they apply? (a) The position of the CM of the rod is X = x₁ + x₂/2, Y = y₁ + y₂/2. For the rod we have: . T is the kinetic energy of the motion of the CM and the motion about the CM. U = -mg(y₁ +y₂/2) + constant. For small oscillations x₁ = lsinq₁ @ lq₁, x₂ = (3/2)lsinq @ (3/2)lq. y₁ = lcosq₁ @ l(1-q₁²/2), y₂ = (3/2)lcosq @ (3/2)l(1-q²/2). To second order in the small quantities we have: T = (1/2)ml²((dq₁/dt)²+(3/4)²(dq/dt)²+(3/2)(dq₁/dt)(dq/dt)+(1/2)(3/16)ml²(dq/dt)²) = (1/2)ml²((dq₁/dt)²+(3/4)(dq/dt)²+(3/2)(dq₁/dt)(dq/dt)) = (1/2)ST_ijq_iq_j. U = mgl[(q₁²/2) +(3/4)l(q²/2)] = (1/2)Sk_ijq_iq_j. q₁ = q₁, q₂ = q, T₁₁ = ml², T₂₂ = (3/4)ml², T₁₂ = T₂₁ = (3/4)ml². k₁₁ = mgl, k₂₂ = (3/4)mgl, k₁₂ = k₂₁ = 0. Details of the calculation: . We have a solution if . (3/4)(g-w²l)² -(9/16)w⁴l² = 0, w⁴ - 8(g/l)w² + 4g²/l = 0. w² = [4 ± (12)^1/2]g/l. mode 1: w₁² = [4 + (12)^1/2]g/l. (g - [4 + (12)^1/2]g)A₁ - (3/4)[4 + (12)^1/2]gA₂ = 0. A₁ = -0.866 A₂. mode 2: w₂² = [4 - (12)^1/2]g/l. (g - [4 - (12)^1/2]g)A₁ - (3/4)[4 - (12)^1/2]gA₂ = 0. A₁ = 0.866 A₂. (b) The most general solution is . q₁ = Re(c₁exp(iw₁t) + c₂exp(iw₂t)), q = (1/0.866)Re(-c₁exp(iw₁t) + c₂exp(iw₂t)). Initial conditions: q₁(0) = d/l, q(0) = 0, dq₁/dt = dq/dt = 0 at t = 0. Re(c₁ + c₂) = d/l, Re(c₁ - c₂) = 0, Re(c₁) = Re(c₂) = d/(2l). Re(iw₁c₁ + iwc₂) = 0, Re(-iw₁c₁ + iwc₂) = 0. Im(w₁c₁ + wc₂) = 0, Im(w₁c₁ - wc₂) = 0, Im(c₁) = Im(c₂) = 0. q₁(t) = (d/(2l))[cosw₁t + cosw₂t], q(t) = (d/(20.866l))[cosw₁t - cosw₂t].

Problem 3:

Two masses a and b are on a horizontal surface. Mass b has a spring connected to it and is at rest. Mass a has an initial velocity v₀ along the x-axis and strikes the spring of constant k, compressing it and thus starting mass b in motion along the x-axis.

(a) Find the maximum force exerted.
(b) Find the resulting motion of mass a while in contact with the spring.

Solution:

	Concepts, principles, relations that apply to the problem: Small oscillations, coupled oscillations, normal modes Solutions of the form q_j = Re(A_jeⁱ^w^t) can be found. We can find the w² from det(k_ij- w²T_ij) = 0.
	Why do they apply? During the time interval that both masses are in contact with the spring, they can be treated as coupled harmonic oscillators.
	How do they apply? (a) Assume that at t = 0 mass a makes contact with the spring. At t = 0 both masses are at their equilibrium position. Let x_a(t) and x_b(t) denote the their displacements from their equilibrium positions, respectively. In the collision energy and momentum are conserved. T_i = T_f + U, P_i = P_f. The maximum force is exerted when the spring has maximum compression. At that instant both masses move with the same velocity. (1/2)m_av₀² = (1/2)(m_a + m_b)v² + (1/2)k(x_b - x_a)_min². m_av₀ = (m_a + m_b)v. Combining the two equations we have (x_b - x_a)_min = (m_am_b/k)^1/2 v₀/(m_a + m_b)^1/2. The maximum force is k(x_b - x_a)_min = (km_am_b)^1/2 v₀/(m_a + m_b)^1/2. (b) T = (1/2)m_a(dx_a/dt)² + (1/2)m_b(dx_b/dt)², U = (1/2)k(x_b - x_a)² = (1/2)k(x_b² + x_a² - x_ax_b - x_bx_a). T₁₁ = m_a, T₂₂ = m_b, k₁₁ = k₂₂ = k, k₁₂ = k₂₁ = -k. det(k_ij- w²T_ij) = 0 --> w₁ = 0, w₂² = k(m_a + m_b)/(m_am_b). For w₁ = 0 we have A₁ = A₂. For w₂² = k(m_a + m_b)/(m_am_b) we have A₂ = -(m_a/m_b)A₁. The most general solutions are: x_a = A₁t + B₁cos(wt + f), x_b = A₁t - (m_a/m_b)B₁cos(wt + f), dx_a/dt = A₁ - wB₁sin(wt + f), dx_b/dt = A₁ + w(m_a/m_a)B₁sin(wt + f).
	Details of the calculations: The initial conditions at t = 0 are x_a = x_b = 0, dx_a/dt = v₀, dx_b/dt = 0. This implies B₁cos(f) = 0, f = p/2. v₀ = A₁ - wB₁, A₁ + w(m_a/m_b)B₁ = 0. A₁ = v₀m_a/(m_a + m_b), B₁ = -(1/w)v₀m_b/(m_a + m_b). Therefore x_a = (v₀/w)m_b/(m_a + m_b)sinwt + m_av₀/(m_a + m_b)t, dx_a/dt = v₀m_b/(m_a + m_b)coswt + m_av₀/(m_a + m_b), as long as mass a is in contact with the spring. [At t = 0 we have dx_a(lab) /dt = v₀m_b/(m_a + m_b) + m_av₀/(m_a + m_b) = v₀.]

A different approach:

Solution:

	Concepts, principles, relations that apply to the problem: Simple harmonic motion, frame transformations, two interacting particles
	Why do they apply? In the center of mass frame we have an elastic collision between two particles with an interaction energy U(x) = (1/2)k(x-x₀)², where x-x₀ is the distance between the particles, as long as particle a is in contact with the spring .
	How do they apply? (a) In the center of mass frame we have an elastic collision. Assume mass a approaches the CM with with speed v_a from the left, and mass b approaches the CM with with speed v_b from the right. The CM is at rest and mass a recedes from the CM with with speed v_a to the right, while mass b recedes from the CM with with speed v_b to the left after the collision. We have m_av_a = m_bv_b in the CM frame before the collision. To find the relationship between the speeds in the CM and the lab frame we use m_a(v₀ - v_b) = m_bv_b. v_b = m_av₀/(m_a + m_b)), v_a = m_bv₀/(m_a + m_b). The maximum force is exerted when all the kinetic energy is converted into potential energy. F_max = k(x_min-x₀). k(x_min-x₀)²= m_av_a² + m_bv_b² = m_a(m_bv₀/(m_a + m_b))² + m_b(m_av₀/(m_a + m_b))² = m_am_bv₀²/(m_a + m_b). k(x_min-x₀) = (km_am_b)^1/2 v₀/(m_a + m_b)^1/2.
	Details of the calculations: (b) Assume that mass a makes contact with the spring at t = 0, at the origin of the CM and the lab frame. (We assume that at t = 0 the origin of the two frames coincide.) Assume that at t = 0 the distance between mass a and mass b is x₀. In the CM frame when mass a moves a distance x_a to the right, mass b moves a distance x_b = x_a(m_a/m_b) to the left. The distance between the masses becomes x = x₀ - (x_a + x_b) = x₀ - x_a(m_a + m_b)/m_b = x₀ - x'. For t > 0 we have U = (1/2)kx^'2 = (1/2)[k(m_a + m_b)²/m_b²]x_a² until the time when contact is lost again. The kinetic energy of the particles is T = (1/2)m_a(dx_a/dt)² + (1/2)m_b(dx_b/dt)² = (1/2)(m_a + m_a²/m_b)(dx_a/dt)². L = T - U = (1/2)(m_a(m_b + m_a)/m_b)(dx_a/dt)² - 1/2)[k(m_a + m_b)²/m_b²]x_a². This is a Lagrangian for simple harmonic motion. d²x_a/dt² = -k(m_b + m_a)/(m_am_b)x_a = -(k/m)x_a, with m = m_am_b/(m_b + m_a). We therefore have x_a = Asinwt, w = (k/m)^1/2, until t = p/w (1/2 period of the motion). dx_a/dt = wAcoswt. At t = 0 we have wA = v_a = m_bv₀/(m_a + m_b), A = m_bv₀/[w(m_a + m_b)] In the lab frame we have x_a(lab) = x_a + v_bt. The motion of mass a is described by x_a(lab) = (v₀/w)m_b/(m_a + m_b)sinwt + m_av₀/(m_a + m_b)t, dx_a(lab) /dt = v₀m_b/(m_a + m_b)coswt + m_av₀/(m_a + m_b), as long as mass a is in contact with the spring. [At t = 0 we have dx_a(lab) /dt = v₀m_b/(m_a + m_b) + m_av₀/(m_a + m_b) = v₀.]