A 1.00 x 10-2 kg block is resting on a horizontal frictionless surface and is attached to ahorizontal spring whose spring constant is 124 N/m. The block is shoved parallel to the springaxis and is given an initial speed of 8.00 m/s, while the spring is initially unstrained. What isthe amplitude of the resulting simple harmonic motion?
Question
A 1.00 x 10-2 kg block is resting on a horizontal frictionless surface and is attached to ahorizontal spring whose spring constant is 124 N/m. The block is shoved parallel to the springaxis and is given an initial speed of 8.00 m/s, while the spring is initially unstrained. What isthe amplitude of the resulting simple harmonic motion?
Solution
To find the amplitude of the resulting simple harmonic motion, we can use the equation for the potential energy stored in a spring:
Potential energy (PE) = (1/2)kx^2
Where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the block is given an initial speed of 8.00 m/s, so it will compress the spring until it comes to a stop and then start moving back in the opposite direction. At the point where the block comes to a stop, all of its initial kinetic energy will be converted into potential energy stored in the spring.
The initial kinetic energy (KE) of the block can be calculated using the equation:
Kinetic energy (KE) = (1/2)mv^2
Where m is the mass of the block and v is its initial velocity.
Given that the mass of the block is 1.00 x 10^-2 kg and its initial velocity is 8.00 m/s, we can calculate the initial kinetic energy:
KE = (1/2)(1.00 x 10^-2 kg)(8.00 m/s)^2 = 0.32 J
Since the initial kinetic energy is equal to the potential energy stored in the spring at the point where the block comes to a stop, we can equate the two:
PE = KE (1/2)kx^2 = 0.32 J
Rearranging the equation, we can solve for x:
x^2 = (2)(0.32 J) / k x^2 = 0.64 J / 124 N/m x^2 = 0.005161 J/N
Taking the square root of both sides, we find:
x = √(0.005161 J/N) x ≈ 0.072 m
Therefore, the amplitude of the resulting simple harmonic motion is approximately 0.072 meters.
Similar Questions
A 500 g block on a frictionless, horizontal surface is attached to a spring with constant k =8.7 N/m. A second block rests on the first, and the whole system executes simple harmonicmotion with period 1.8 s. When the amplitude of the motion is increased to 35 cm the upperblock just begins to slip. Determine the coefficient of static friction between the blocks
A 2.0 kg mass is connected to a spring with a spring constant of 9.0 N/m. The displacement is given by the expression x(t) = (12 cm) sin(ω t). What is the amplitude of the motion?
A 0.720 kg wooden block is attached vertically to a spring with a spring constant of 24.0 N/m. The wooden block is pulled down 8.40 cm from rest and then released. What is the time constant of the wooden block if its amplitude of oscillation is decreased to 5.00 cm after 28.0 oscillations? 49.6 s 29.4 s 10.2 s 6.15 s
wo springs with negligible masses and force constants k1=200 N/m𝑘1=200 N/m and k2=160 N/m𝑘2=160 N/m are attached to the block of mass m=10 kg𝑚=10 kg as shown in figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t=0,𝑡=0, sharp impulse of 50 N−s50 N-s is given to the block with a hammer along the spring. Then the amplitude of spring will be......m......m.
Two blocks with masses m1=1kg and m2=2 kg are connected by a spring constant k=24 Nm−1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0=12 cms −1 to the right, the amplitude of oscillation 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.