Two equal masses are attached to separate identical springs next to one another. Onemass is pulled so its spring stretches 20 cm and the other is pulled so its springstretches only 10 cm. The masses are released simultaneously. Which mass reachesthe equilibrium point first?
Question
Two equal masses are attached to separate identical springs next to one another. Onemass is pulled so its spring stretches 20 cm and the other is pulled so its springstretches only 10 cm. The masses are released simultaneously. Which mass reachesthe equilibrium point first?
Solution
The mass that is stretched only 10 cm will reach the equilibrium point first. Here's why:
-
The time it takes for a mass-spring system to complete one cycle of oscillation (i.e., to return to its equilibrium position) is determined by its period, which is given by the formula T = 2π√(m/k), where m is the mass and k is the spring constant.
-
In this case, the two masses (m) are equal and the two springs are identical, so their spring constants (k) are also equal. Therefore, the periods of the two systems are the same.
-
However, the mass that is stretched 20 cm has to cover a greater distance to return to its equilibrium position than the mass that is stretched 10 cm.
-
Since they both oscillate at the same rate (i.e., they have the same period), the mass that has less distance to cover (the one stretched 10 cm) will reach its equilibrium position first.
Similar Questions
An ideal spring is connected to the ceiling as shown above. A mass M=7kg is connected to the bottom end of the spring and is allowed to reach a new equilibrium position, stretching the spring a displacement L=0.075m. The spring is then raised from the equilibrium position to a height ymax = 0.035 m and released. What is the position of the mass relative to equilibrium at t=1.1 (treat upward as the +y direction)?
Two light springs (1) and (2) of spring constant k and 2k are connected in series as shown in figure. A block of mass M is hanging vertically in equilibrium. If Δx is the elongation in spring (2) in cm, then value of Δx will be (Given, Mgk = 8 cm)Given, 𝑀𝑔𝑘 = 8 cm
Spring calculations: A spring stretches 0.150n m when a 0.3 kgmass is gently suspended from it. The spring is then set uphorizontally with the 0.3 kg mass resting on the frictionless table.The mass is pulled so that the spring is stretched 0.1 m from theequilibrium point and released from rest. Determine a) the springstiffness constant k, b) the amplitude of the maximum acceleration.
A system consist of two springs and a mass m=1 kg 𝑚=1 kg as shown in figure. If mass m is displaced slightly along vertical and released, the system oscillates with a period of 2s. Then the spring constant k is
Two blocks each of the mass 𝑚 areconnected to a spring of spring constant𝑘. If both are given velocity 𝜐 in oppositedirections, then the maximumelongation of the spring is
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.