Motion Test - 8

As minimum velocity in SHM occurs at centre mean position, ∴ KE at centre = Total energy - P.E at centre = 9 - 5 = 4 joules.

K.E =1/2M V²

4=1/2.2(V²)

V²=4,V=2m/s

In SHM, V_man=AW,A=0.01m

2 = 0.01 × W,W = 200 

Time period (T) =$\frac{2\mathrm{\pi }}{w}=\frac{2\mathrm{\pi }}{200}=\frac{\mathrm{\pi }}{100}sec$

According to the question the correct option will be ‘D’ because If the particle moves with uniform speed in a circle, the projection of its path along its diameter is an oscillatory motion. Thus option (d) is the correct option.

According to the question, the correct option will be ‘A’ because when the bus suddenly starts moving in the forward direction. This happens due to. When a bus suddenly starts, the standing passengers fall backward on the bus.
It is due to the inertia of rest.

According to the question the correct option will be ‘C’ because Circular motion is the motion of an object that moves at a fixed distance from a fixed point. Here all objects rotate in circular motion.

According to the question the correct option will be ‘C’ because Acceleration is proportional to displacement from mean, thus it changes.
Velocity also changes as it is zero at extreme and maximum at centre.
Phase also changes, but for the oscillations to be undammed, it must go to the same extreme position each time, or amplitude must be constant.

According to the question the correct option will be ‘B’ because 

No of oscillation ,n = $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{B\sin \theta }{i}}$

$\frac{20}{15}=\sqrt{\frac{B_{1}sin30}{B_{2}sin60}}$

$\frac{B_1\left(\frac{1}{2}\right)}{B_2\left(\frac{\sqrt{3}}{2}\right)}=\frac{{20}^2\sqrt{3}}{{15}^2}$

=$1.7\sqrt{3}$ 

Hence, Option B is correct answer.

\(a=\omega^{2} x\)

\(\mathrm{S} \mathrm{o}, a_{m a x}=\omega^{2} A\)

We know that \(\omega=2 \pi f\)

\(\mathrm{So}, a_{m a x}=\frac{(2 \pi \times 2)^{2} \times 5}{100}\)

Case I:

\(N-m g=m a_{m a x}\)

\(N=m\left(a_{m a x}+g\right)\)

\(=60\left(10+\frac{16 \times \pi^{2} \times 5}{100}\right)\)

\(=1080\)

\(N=108 \mathrm{kg}-\mathrm{wt}\)

Case II:

\(m g-N=m a_{m a x}\)

or, \(N=m g-m a_{\text {max }}\)

\(=60(10-8)\)

\(=120 N=12 \mathrm{kg}-\mathrm{wt}\)

According to the question, the correct option will be ‘C’ because Motions of the propeller of a flying helicopter, the minute hand of watch and tape of the cassette recorder are the example of rotatory motion.

According to the question the correct option will be ‘B’ because
Resultant amplitude is A.

Here a₁= 1,a₂= 1

According to the question the correct option will be ‘B’ because the time taken for a particle to a particle to oscillate from one extreme to other extreme is equal to half the time period i.e., $\frac{T}{2}.$

∴$\frac{T}{2}=4s\left(given\right)$

T = 8s

ω =$\frac{2\mathrm{\pi }}{1}=\frac{2\mathrm{\pi }}{8}=0.25\mathrm{\pi }$

According to the question the correct option will be ‘C’ because The time period of motion is related to its angular velocity using the equation T=2π/ωT=2π/ω

Thus, (c) is the correct option.

According to the question the correct option will be ‘B’ because the restoring force tries to bring the body back to its mean position.

Basic theory, \((d x, d y)\) is called as transitional or shift vector.

According to the question the correct option will be ‘B’ because θ = ϕsin(ωt)θ is the equation of angular SHM, θ is the instantaneous angular displacement and ϕ is the angular displacement of the bob

The correct option is (b)

According to the question the correct option will be ‘B’ because the bob undergoes SHM, while the string undergoes angular SHM
the correct option is (b)