Oscillations - Free MCQ Practice Test with solutions, NEET NCERT Based

The bob of a pendulum moves in such a way that it repeats its positions several time but alternately the time gap between these positions is always equal which proves that the motion of a pendulum is oscillatory.

In a simple pendulum the restoring force is due to the tangential component of the gravitational force because on applying torque equation on the radial forces the effect is nullified due to it passing from centre & only tangential component remains.

We know that PE + KE = TE = constant
Hence at the extremis TE = PE = ½ kz²  
Where z is amplitude and k is shm constant.
Thus when KE = ⅓ PE
We get PE = ¾ TE =  ½ kz²
Hence we get  ½ kx² = ¾  ½ kz²
We get x/z = √3/4 
= 1.73 / 2
= .87
Thus we get x is 87 percent of the amplitude.

Time period, T=2π√l/g
=>T∝√l
=>ΔT/T=1/2 Δl/l
Δl/l=1%
Therefore, ΔT/T=1/2x1%=0.5%

v=ω√A²−x2
​⇒8=ω √A²−25​
&10=ω√A²−16
​⇒(8/10​)2= A²−25/ A²−16
​⇒16/25​= A²−25/ A²−16
​⇒16A²−256=25A²−625
⇒9A²=369
A²=41
⇒8=ω41−25
​⇒8=4ω
∴ω=2=T2π​
T=π sec​

Oscillatory motion is periodic to and fro motion. By definition

Maximum acceleration = w2A
Maximum velocity = wA
Ratio = w2A/wA = w
w = angular velocity
A = amplitude

Damping is due to resistive forces like air drag, friction etc.

T1=2π√1/g=T, T₂=2π√16/g=4T
(ω1−ω2)t=2π
(2π/T−2π/4T)t=2π
t=4T/3

Explanation:Time period of an oscillatory motion is the smallest interval of time for one complete oscillation.As when we apply the force on the bob,bob moves from mean position to extreme position then form extreme position to mean position and then again from mean to extreme in opposite direction and at last come to mean position.The time taken by bob in this whole process is called period of oscillation.

For all practical purposes, the motion of a simple pendulum is SHM only if the maximum angle which the string makes with the vertical is less than 1 because for angles less than 1° sinθ = θ.

Simple pendulum is not a case of ideal SHM as it is mention in NCERT.
Shadow of a point mass having uniform circular motion on the horizontal diameter is a case of SHM but not on the vertical diameter.

T=2*3.14*√m/√k=1 s (given)

T’=2*3.14*√4m/√k=2*3.14*2√m/√k=2*1s

Therefore T’=2s

If a resonant mechanical structure is set in motion and left to its own devices, it will continue to oscillate at a particular frequency known as its natural frequency, or "damped natural frequency". This will be a little lower in frequency than the resonant frequency, which is the frequency it would assume if there were no damping. The resonant frequency is also called the "undamped natural frequency”

When rubber ball completely filled with water its centre of gravity will be at its centre, as water will fall through hole its COG will shift towards lower side leading to increase in length of pendulum and thus T, when very small amount of water will be left in rubber ball its COG will again shifts upward causing decrease in length and thus T, and finally when rubber ball becomes empty its COG will be at its centre and T will remains same as earlier.

Time period remains same since
T = 2π √(L/g)
So, from proportionality we can solve

All periodic functions can, in principle, be represented as a sum of sine and cosine functions, since they form what's known as a complete set.

L’=1.001L
T=2π√L/9=1s
T’=2π√L’/g=2π√1.001/g=Tx√1.001=1.0005xT
T’=1.0005s
Error in time per day=1.0005x60x60x24-1x60x60x24
Error=86443.2-86400
Error=43.2s