Question

A mass \[m = 100gms\] is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to \[0.16meter\] and the time period equal to\[2\sec \]. Initially the mass is released from rest at \[t = 0\] and displacement\[x = - 0.16meter\]. The expression for the displacement of the mass at any time (t) is
(a) \[x = 0.16\cos \left( {\pi t} \right)\]
(b) \[x = - 0.16\cos \left( {\pi t} \right)\]
(c) \[x = 0.16\cos \left( {\pi t + \pi } \right)\]
(d) \[x = - 0.16\cos \left( {\pi t + \pi } \right)\]

Answer 1

Hint: From the given data, we identified that the spring-mass system formula will be used to derive the expression for the displacement of the mass at any time. The values of amplitude and time period of a light spring that oscillates on a frictionless horizontal table are given in the question. So, we will apply these values directly in the spring-mass system formula.

Complete step-by-step solution:
For spring-mass system
\[x = a\cos \omega t\]
We know that, the angular velocity \[\omega = \dfrac{{2\pi }}{T}\]
\[x = a\cos \dfrac{{2\pi }}{T}t\]
Where, \[a\]-amplitude, \[T - \]Time period
Given that, \[a = - 0.16m\]and \[T = 2\sec \]
The displacement \[\left( x \right)\] of the mass at any time \[\left( t \right)\] is
 \[x = - 0.16\cos \dfrac{{2\pi }}{2}t\]
\[x = - 0.16\cos \pi t\]
Hence, the correct option is B.

Additional Information: In a spring-mass system, there’ll not be any external forces performing on them. A system of masses connected by springs may be a classical system with several degrees of freedom. A system consisting of two masses and three springs has two degrees of freedom. Because the energy is conserved in any spring-mass system, the speed of the block at any point is easily calculated using the energy conservation system. The motion of a mass on a spring will be described as simple harmonic motion and oscillatory motion that follows Hooke’s law. The natural frequency is additionally called Eigen frequency, which is the frequency at which a system tends to oscillate within the absence of any driving or damping force. The motion pattern of a system oscillating at natural frequency is termed normal mode.

Note: For clarity, this doesn’t imply that the acceleration does not affect a spring-mass system; it just means it’s no effect on the frequency of the system. The acceleration has a sway on the equilibrium position of the system. For instance, if the system wasn’t oscillating and is in equilibrium, the position of the mass will shift within the direction opposite that of the acceleration.