A 10 kg metal block is attached to a spring of spring constant 10
A. 10 ms<sup>-2</sup>
B. 100 ms<sup>-2</sup>
C. 200 ms<sup>-2</sup>
D. 0.1 ms<sup>-2</sup>
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
CONCEPT:
- Simple harmonic motion: A periodic motion in which the restoring force on the object is opposite in direction and directly proportional to the object's displacement.
Restoring force (F) = - k x
Where k is a constant and x is the displacement from the mean position.
- Amplitude: The maximum displacement of the object from equilibrium, either in the positive or negative x-direction.
- Time Period: The time in which the object completes one oscillation called time period.
The angular frequency ω in terms of the time period is given by:
ω = 2π/T
where ω is the angular frequency and T is the time period.
For spring-mass SHM, angular frequency ω of the object is also given by:
ω = \(\sqrt{\frac{K}{m}}\)
where ω is the angular frequency, m is the mass of the object and K is the spring constant.
- Acceleration: Acceleration on the moving object is directly proportional to the object's displacement vector and the opposite of displacement in direction.
The maximum value of acceleration is given by:
amax = Aω2
where A is the amplitude of oscillation and ω is the angular frequency.
CALCULATION:
Given that m = 10 kg; K = 1000 N/m; A = 10cm = 0.1m
ω = \(\sqrt{\frac{K}{m}} = \sqrt{\frac{1000}{10}}\)
ω = 10 radian per sec
amax = Aω2
amax = 0.1 × 102
amax = 10 m/s2
So the correct answer is option 1.