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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
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Q.1
A particle is executing SHM at midpoint of mean position and extreme position . What is it's KE in terms of total energy E.
0%
E/2
0%
4E/3
0%
√ 2 E
0%
3E/4
Explanation
Solution:
Potential energy at displacement x is PE=½kx². At the midpoint between the mean position and the extreme, x=A/2:
PE = ½k(A/2)² = ⅛kA² = ¼(½kA²) = E/4.
KE = E - PE = E - E/4 = 3E/4.
Q.2
Total energy of mass spring system in harmonic motion is E=1/2(mω2A2). Consider another system executing SHM with same amplitude having value of spring constant as half the previous one and mass twice as that of previous one. The energy of second oscillator will be
0%
E
0%
2E
0%
$\sqrt 2 E$
0%
E/2
Explanation
Solution:
For SHM, total energy also equals ½kA² (the same E, written using the spring constant instead of ω and m). Halving k while keeping amplitude A the same directly halves the energy: E' = ½(k/2)A² = ½E, regardless of the mass.
Q.3
A solid cylinder of radius r and mass m is connected to a spring of spring constant k and it slips on a friction less surface without rolling with angular frequency
0%
√(k/mr)
0%
√(kr/m)
0%
√(k/m)
0%
√(2k/m)
Explanation
Solution:
The cylinder here slips freely without rolling, so no rotational inertia is involved at all -- it's just an ordinary block-and-spring system, oscillating at the standard ω=√(k/m).
Q.4
The total energy of the particle executing SHM is Here x is the displacement of the particle
0%
$ \alpha x^2$
0%
$ \alpha x$
0%
$ \alpha x^{1/2}$
0%
Independent of x
Explanation
Solution:
The total mechanical energy of an SHM oscillator, kinetic plus potential, stays fixed at ½kA² throughout the motion -- it doesn't depend on the instantaneous displacement x at all, even though KE and PE individually trade off against each other as x changes.
Q.5
In a spring mass system executing SHM having mass m and spring constant K with Time period T. Match the column I to the column II
0%
p-> iv, q -> i ,r -> iii, s-> ii
0%
p-> iii, q -> i ,r -> iv, s-> ii
0%
p-> i, q -> i ,r -> iii, s-> ii
0%
p-> iii, q -> i ,r -> iii, s-> ii
Explanation
Solution:
T=2π√(m/k) depends only on mass and spring constant -- never on amplitude.
p) Amplitude doubled (nothing else changes): T is unaffected -- Time period is T.
q) Mass doubled (k unchanged): T'=2π√(2m/k)=√2·T -- Time period is T√2.
r) Both k and m doubled: T'=2π√(2m/2k)=2π√(m/k)=T -- the two changes cancel, Time period is T.
s) Amplitude doubled (irrelevant to T) and spring constant halved: with only k→k/2 actually affecting the period, T'=2π√(m/(k/2)) =√2·T, the same T√2 result as row q -- a softer spring means a slower, longer-period oscillation, not a shorter one.
Q.6
If a simple harmonic motion is represented by $\frac {d^2x}{dt^2} + \beta x=0$, then its time period is
0%
$\frac {2 \pi}{\sqrt {\beta}}$
0%
$\frac {2 \pi}{\beta}$
0%
$2 \pi \beta$
0%
$2 \pi\sqrt {\beta}$
Explanation
Solution:
Comparing d²x/dt²+βx=0 to the standard SHM equation d²x/dt²=-ω²x shows ω²=β, so ω=√β, and T=2π/ω=2π/√β.
Q.7
A particle is executing linear SHM of amplitude A. What fraction of total energy is potential when the displacement is 1/4 times amplitude
0%
3/2
0%
1/16
0%
1/4
0%
1/2√ 2
Explanation
Solution:
The potential energy fraction at displacement x is (½kx²)/(½kA²) = x²/A². At x=A/4: (A/4)²/A² = (A²/16)/A² = 1/16.
Q.8
In the system shown below frequency of oscillation when mass is displaced slightly is
0%
$f=\frac {1}{2\pi} \sqrt { \frac {k_1k_2}{k_1+k_2)m}}$
0%
$f=\frac {1}{2\pi} \sqrt { \frac {k_1+k_2}{m}}$
0%
$f=\frac {1}{2\pi} \sqrt { \frac {m}{k_1 k_2}}$
0%
$f=\frac {1}{2\pi} \sqrt { \frac {k_1+k_2}{mk_1 k_2}}$
Explanation
Solution:
The two springs k
1
and k
2
are connected end to end between the wall and the mass -- a series combination, whose effective spring constant is k
eff
=k
1
k
2
/(k
1
+k
2
) (they combine like resistors in parallel).
f = (1/2π)√(k
eff
/m) = (1/2π)√[k
1
k
2
/((k
1
+k
2
)m)].
Q.9
find the maximum static frictional force on the body M2
0%
$\mu _s g M_1$
0%
$\mu _s g M_2$
0%
$\mu _s g (M_1 + M_2)$
0%
None of these
Explanation
Solution:
M
2
sits on top of M
1
with no vertical acceleration, so the normal force between them just supports M
2
's own weight, M
2
g. The maximum static friction available is that normal force times the friction coefficient: μ
s
gM
2
.
Q.10
A solid cylinder of mass m is attached to a horizontal spring with force constant k. The cylinder can roll without slipping along the horizontal plane. Center of mass of the cylinder executes simple harmonic motion if displaced from its mean position,The time period is
0%
$T= 2 \pi \sqrt { \frac {5m}{k}}$
0%
$T= 2 \pi \sqrt { \frac {3m}{2k}}$
0%
$T= 2 \pi \sqrt { \frac {m}{3k}}$
0%
$T= 2 \pi \sqrt { \frac {m}{k}}$
Explanation
Solution:
For a cylinder rolling without slipping (I=½mr²), total kinetic energy is KE = ½mv²+½Iω² = ½mv²+¼mv² = ¾mv² (using v=ωr).
Total energy ¾mv²+½kx² stays constant; differentiating with respect to time and dividing through by v gives &frac32;m(dv/dt) = -kx, i.e. a = -[2k/(3m)]x -- SHM with ω²=2k/(3m).
T = 2π/ω = 2π√(3m/2k).
Q.11
Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?
0%
When v is maximum, a is maximum
0%
Value of a is zero, whatever may be the value of v.
0%
When v is maximum, a is zero.
0%
When v is zero, a is zero
Explanation
Solution:
In SHM, acceleration is a=-ω²x -- proportional to displacement, so it's zero exactly at the mean position (x=0) and maximum at the extremes. Velocity is the opposite: it's maximum at the mean position and zero at the extremes. So whenever v is at its maximum (at the mean position), a is zero there.
Q.12
Find the maximum oscillation magnitude Amax that permits the two bodies to move as unit
0%
$A_{max}=\frac {(1+a)\mu _s gM_1}{k}$
0%
$A_{max}=\frac {\mu _s gM_1}{k}$
0%
$A_{max}=\frac {\mu _s gM_2}{k}$
0%
none of the above
Explanation
Solution:
Treating M
1
+M
2
as one oscillating mass, ω²=k/(M
1
+M
2
) = k/(M
1
(1+a)) (using M
2
=aM
1
).
The two bodies stay together as long as the friction needed to accelerate M
2
at the system's peak acceleration, ω²A, doesn't exceed the maximum available friction from Q217, μ
s
gM
2
:
M
2
ω²A
max
= μ
s
gM
2
⇒ A
max
= μ
s
g/ω² = μ
s
gM
1
(1+a)/k.
Q.13
A simple pendulum is displaced from its mean position on to a position A such that height of A above O is 0.05m. It is then released its velocity when it passes mean position is
0%
.1m/s
0%
5.0m/s
0%
1m/s
0%
1.5m/s
Explanation
Solution:
Energy conservation from the release point down to the mean position: ½mv² = mgh ⇒ v = √(2gh).
v = √(2 × 10 × 0.05) = √1 = 1 m/s.
Q.14
The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be
0%
T
0%
2T
0%
T/2
0%
T/4
Explanation
Solution:
Cutting a spring into n equal pieces multiplies each piece's spring constant by n (a shorter spring of the same material is stiffer) -- cutting into 4 pieces gives each piece a constant of 4k.
With the same mass on one such piece: T' = 2π√(m/(4k)) = ½×2π√(m/k) = T/2.
Q.15
A spring of force constant k is cut into two pieces such that one piece is four times the length of the other. the longer piece will have force constant equal to
0%
4k/5
0%
5k/4
0%
3k/2
0%
4k
Explanation
Solution:
A spring's constant is inversely proportional to its length -- cutting off a piece of length l from a spring of length L and constant k gives that piece a constant of k(L/l).
If one piece is 4× the length of the other, the total length splits into L/5 (short piece) and 4L/5 (long piece). The longer piece's constant is k×L/(4L/5) = k×5/4 = 5k/4.
Q.16
Fig below shows two spring mass systems. All the springs are identical having spring constant k and are of negligible mass. If m is the mass of block attached to the spring then the ratio of time period of oscillations of both systems is
0%
1:2√ 2
0%
2√ 2:1
0%
1:√ 2
0%
√ 2:1
Explanation
Solution:
System (a): one spring k holding mass m: T
a
=2π√(m/k).
System (b): two equal springs k connected in series before the mass, giving an effective constant k
eff
=k²/(2k)=k/2, so T
b
=2π√(m/(k/2))=√2·T
a
.
Ratio T
a
:T
b
= 1:√2.
Q.17
Fig below shows two equal masses of mass m joined by a rope passing over a light pulley. First mass is attached to a spring and another end of spring is attached to a rigid support. Neglecting frictional forces total energy of the system when spring is extended by a distance x is Here, v = dx/dt , the velocity of mass
0%
mv2+1/2(Kx2)+mgx
0%
mv2-1/2(Kx2)+mgx
0%
mv2-1/2(Kx2)-mgx
0%
mv2+1/2(Kx2)-mgx
Explanation
Solution:
Both masses move together at the same speed v (linked by the inextensible rope over the pulley), so their combined kinetic energy is ½mv²+½mv²=mv².
The spring stores ½Kx² of elastic PE as it stretches by x.
As the hanging mass descends by that same x (pulled down as the spring stretches), it LOSES gravitational PE, contributing -mgx to the total.
Total energy = mv² + ½Kx² - mgx.
Q.18
A block of mass M1 resting on the frictionless surface is connected to a spring of spring constant k that is anchored in the nearby wall. A block of mass M2=aM1 is placed on the top of the first block. The coefficient of static friction between the two bodies is μs Assuming the two bodies moves as single object, find the period of oscillation of the system
0%
$T=2 \pi \sqrt {\frac {(1+a)M_1}{k}}$
0%
$T=2 \pi \sqrt {\frac {M_1}{k}}$
0%
$T=2 \pi \sqrt {\frac {(1-a)M_1}{k}}$
0%
$T=2 \pi \sqrt {\frac {aM_1}{k}}$
Explanation
Solution:
Treating M
1
+M
2
as a single oscillating mass (M
2
=aM
1
, so combined mass = M
1
(1+a)):
T = 2π√[(M
1
+M
2
)/k] = 2π√[(1+a)M
1
/k].
0 h : 0 m : 1 s
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