M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip

Questions

M.C.Q (1 Marks)

Take a timed test

28 questions · auto-graded multiple-choice test.

MCQ 11 Mark

A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is $T$. If the spring is divided in two equal parts and one part is used to continue the simple harmonic motion, the time period will:

A
Remain $T$
B
Become $2T$
C
Become $\frac{\text{T}}{2}$
✓
Become $\frac{\text{T}}{\sqrt{2}}$

Answer

Correct option: D.

Become $\frac{\text{T}}{\sqrt{2}}$

Time period $(T)$ is given by,
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
where $m$ is the mass, and $k$ is spring constant.
When the spring is divided into two parts, the new spring constant $k_1$ is given as,
$\text{k}_1=2\text{k}$
New time period $T_1$:
$\text{T}_1=2\pi\sqrt{\frac{\text{m}}{2\text{k}}}=\frac{1}{\sqrt{2}}2\pi\sqrt{\frac{\text{m}}{\text{k}}}=\frac{1}{\sqrt{2}}\text{T}$

View full question & answer→

MCQ 21 Mark

The total mechanical energy of a spring $-$ mass system in $1$ simple harmonic motion is $\text{E}=\frac{1}{2}\text{m}\omega^2\text{A}^2.$ Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

A
Become $2E$
B
Become $\frac{\text{E}}{2}$
C
Become $\sqrt{2\text{E}}$
✓
Remain $E$.

Answer

Correct option: D.

Remain $E$.

Mechanical energy $(E)$ of a spring $-$ mass system in simple harmonic motion is given by,
$\text{E}=\frac{1}{2}\text{m}\omega^2\text{A}^2$
where $m$ is mass of body, and $\omega$ is angular frequency.
Let $m_1$ be the mass of the other particle and $\omega_1$ be its angular frequency.
New angular frequency $\omega_1$ is given by,
$\omega_1=\sqrt{\frac{\text{k}}{\text{m}_1}}=\sqrt{\frac{\text{k}}{2\text{m}}}\big(\text{m}_1=2\text{m}\big)$
New energy $E_1$ is given as,
$\text{E}_1=\frac{1}{2}\text{m}_1\omega_1^2\text{A}^2$
$=\frac{1}{2}\big(2\text{m}\big)\Big(\sqrt{\frac{\text{k}}{2\text{m}}}\Big)^2\text{A}^2$
$=\frac{1}{2}\text{m}\omega^2\text{A}^2=\text{E}$

View full question & answer→

MCQ 31 Mark

A particle moves in a circular path with a uniform speed. Its motion is:

✓
Periodic.
B
Oscillatory.
C
Simple harmonic.
D
Angular simple harmonic.

Answer

Correct option: A.

Periodic.

Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position.

View full question & answer→

MCQ 41 Mark

A particle is fastened at the end of a string and is whirled in a vertical circle with the other end of the string being fixed. The motion of the particle is:

✓
Periodic.
B
Oscillatory.
C
Simple harmonic.
D
Angular simple harmonic.

Answer

Correct option: A.

Periodic.

Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position.

View full question & answer→

MCQ 51 Mark

A pendulum clock that keeps correct time on the earth is taken to the moon. It will run:

A
At correct rate.
B
6 times faster.
C
$\sqrt{6}$ times faster.
✓
$\sqrt{6}$ times slower.

Answer

Correct option: D.

$\sqrt{6}$ times slower.

The acceleration due to gravity at moon is $\frac{\text{g}}{6}. $
Time period of pendulum is given by, $\text{T}=2\pi\sqrt{\frac{\text{l}}{g}}$
Therefore, on moon, time period will be:
$\text{T}_{\text{moon}}=2\pi\sqrt{\frac{\text{l}}{\text{g}_{\text{moon}}}}=2\pi\sqrt{\frac{\text{l}}{\Big(\frac{\text{g}}{6}\Big)}}$
$=\sqrt{6}\Big(2\pi\sqrt{\frac{\text{l}}{\text{g}}}\Big)=\sqrt{6}\text{T}$

View full question & answer→

MCQ 61 Mark

The average energy in one time period in simple harmonic motion is:

✓
$\frac{1}{2}\text{m}\omega^2\text{A}^2$
B
$\frac{1}{4}\text{m}\omega^2\text{A}^2$
C
$\text{m}\omega^2\text{A}^2$
D
$\text{Zero}$

Answer

Correct option: A.

$\frac{1}{2}\text{m}\omega^2\text{A}^2$

It is the total energy in simple harmonic motion in one time period.

View full question & answer→

MCQ 71 Mark

Two bodies $A$ and $B$ of equal mass are suspended from two separate massless springs of spring constant $k_1$ and $k_2$ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of $A$ to that of $B$ is:

A
$\frac{\text{k}_1}{\text{k}_2}$
B
$\sqrt{\frac{\text{k}_1}{\text{k}_2}}$
C
$\frac{\text{k}_2}{\text{k}_1}$
✓
$\sqrt{\frac{\text{k}_2}{\text{k}^1}}$

Answer

Correct option: D.

$\sqrt{\frac{\text{k}_2}{\text{k}^1}}$

Maximum velocity, $\text{v}=\text{A}\omega$
where $A$ is amplitude and $\omega$ is the angular frequency.
Further, $\omega=\sqrt{\frac{\text{k}}{\text{m}}}$
Let $A$ and $B$ be the amplitudes of particles $A$ and $B$ respectively.
As the maximum velocity of particles are equal,
i.e. $v_A= v_B$
or,
$\text{A}\omega_\text{A}=\text{B}\omega_\text{B}$
$\Rightarrow\text{A}\sqrt{\frac{\text{k}_1}{\text{m}_\text{A}}}=\text{B}\sqrt{\frac{\text{k}_2}{\text{m}_\text{B}}}$
$\Rightarrow\text{A}\sqrt{\frac{\text{k}_1}{\text{m}}}=\text{B}\sqrt{\frac{\text{k}_2}{\text{m}}}\big(\text{m}_\text{A}=\text{m}_{\text{B}}=\text{m}\big)$
$\Rightarrow\frac{\text{A}}{\text{B}}=\sqrt{\frac{\text{k}_2}{\text{k}_1}}$

View full question & answer→

MCQ 81 Mark

The average acceleration in one time period in a simple harmonic motion is:

A
$\text{A}\omega^2$
B
$\frac{\text{A}\omega^2}{2}$
C
$\frac{\text{A}\omega^2}{\sqrt{2}}$
✓
$\text{Zero}.$

Answer

Correct option: D.

$\text{Zero}.$

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
$\text{A}\omega^2+(-\text{A}\omega^2)=0$

View full question & answer→

MCQ 91 Mark

A pendulum clock keeping correct time is taken to high altitudes:

A
It will keep correct time.
B
Its length should be increased to keep correct time.
✓
Its length should be decreased to keep correct time.
D
It cannot keep correct time even if the length is changed.

Answer

Correct option: C.

Its length should be decreased to keep correct time.

Time period of pendulum,
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
At higher altitudes, the value of acceleration due to gravity decreases.
Therefore, the length of the pendulum should be decreased to compensate for the decrease in the value of acceleration due to gravity.

View full question & answer→

MCQ 101 Mark

A particle executes simple harmonic motion with a frequency v. The frequency with which the kinetic energy oscillates is:

A
$\frac{\text{v}}{2}$
B
v
✓
2v
D
zero

Answer

Correct option: C.

2v

Because in one complete oscillation, the kinetic energy changes its value from zero to maximum, twice.

View full question & answer→

MCQ 111 Mark

A particle moves on the $X-$ axis according to the equation $\text{x = A + B}\sin\omega\text{t}.$ The motion is simple harmonic with amplitude:

A
$A$
✓
$B$
C
$A + B$
D
$\sqrt{\text{A}^2+\text{B}^2}$

Answer

Correct option: B.

$B$

At $t = 0,$
Displacement $(x_0)$ is given by,
$\text{x}_0=\text{A}+\sin\omega(0)=\text{A}$
Displacement x will be maximum when $\sin\omega\text{t}$ is $1$
or,
$x_m= A + B$
Amplitude will be:
$x_m-x_0=A+B-A=B$

View full question & answer→

MCQ 121 Mark

A student says that he had applied a force $\text{F}=-\text{k}\sqrt{\text{x}}$ on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.

✓
As x increases k increases.
B
As x increases k decreases.
C
As x increases k remains constant.
D
The motion cannot be simple harmonic.

Answer

Correct option: A.

As x increases k increases.

A body is said to be in simple harmonic motion only when,
F = -kx .....(1)
where F is force,
k is force constant, and,
x is displacement of the body from the mean position.
Given:
$\text{F}=-\text{k}\sqrt{\text{x}}\ ...(2)$
On comparing the equations (1) and (2), it can be said that in order to execute simple harmonic motion, k should be proportional to $\sqrt{\text{x}}$ Thus, as x increases k increases.

View full question & answer→

MCQ 131 Mark

The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is:

A
The mean position.
✓
An extreme position.
C
Between the mean position and the positive extreme.
D
Between the mean position and the negative extreme.

Answer

Correct option: B.

An extreme position.

One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.

View full question & answer→

MCQ 141 Mark

The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity $\overrightarrow{\text{v}}.$ The value of $v$ is

✓
$v_{\max}$
B
$0$
C
between $0$ and $v_{\max}$
D
between $0$ and $- v_{\max}$

Answer

Correct option: A.

$v_{\max}$

Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation.

View full question & answer→

MCQ 151 Mark

A spring-mass system occillates in a car. if the car accelerates on a horizontal road, the frequency of oscillation will:

A
Increase.
B
Decrease.
✓
Remain same.
D
Become zero.

Answer

Correct option: C.

Remain same.

As the frequency of the system is independent of the acceleration of the system.

View full question & answer→

MCQ 161 Mark

A spring-mass system oscillates with a frequency v. If it is taken in an elevator slowly accelerating upward, the frequency will:

A
Increase.
B
Decrease.
✓
Remain same.
D
Become zero.

Answer

Correct option: C.

Remain same.

Because the frequency $\Big(\text{v}-\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}\Big)$ of the system is independent of the acceleration of the system.

View full question & answer→

MCQ 171 Mark

The displacement of a particle is given by $\overrightarrow{\text{r}}=\text{A}\big(\overrightarrow{\text{i}}\cos\omega\text{t}+\overrightarrow{\text{j}}\sin\omega\text{t}\big)$ The motion of the particle is:

A
Simple harmonic.
B
On a straight line.
✓
On a circle.
D
With constant acceleration.

Answer

Correct option: C.

On a circle.

We know,
$\frac{\text{d}^2}{\text{dt}^2}\overrightarrow{\text{r}}=-\omega^2\overrightarrow{\text{r}}$
But there is a phase difference of $90^\circ$ between the $x$ and $y$ components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.

View full question & answer→

MCQ 181 Mark

For a particle executing simple harmonic motion, the acceleration is proportional to:

✓
Displacement from the mean position.
B
Distance from the mean position.
C
Distance travelled since t = 0.
D
Speed.

Answer

Correct option: A.

Displacement from the mean position.

For S.H.M.,
F = -kx
ma = -kx (F = ma)
or,
$\text{a}=-\frac{\text{k}}{\text{m}}\text{x}$
Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.

View full question & answer→

MCQ 191 Mark

The distance moved by a particle in simple harmonic motion in one time period is:

A
A
B
2A
✓
4A
D
zero.

Answer

Correct option: C.

4A

In an oscillation, the particle goes from one extreme position to other extreme position that lies on the other side of mean position and then returns back to the initial extreme position. Thus, total distance moved by particle is,
2A + 2A = 4A.

View full question & answer→

MCQ 201 Mark

A particle moves in a circular path with a continuously increasing speed. Its motion is:

A
Periodic.
B
Oscillatory.
C
Simple harmonic.
✓
None of them.

Answer

Correct option: D.

None of them.

As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position.

View full question & answer→

MCQ 211 Mark

A particle moves on the $X-$ axis according to the equation $\text{x}=\text{x}_0\sin^2\omega\text{t.}$ The motion is simple harmonic.

A
With amplitude $x_0$
B
With amplitude $2x_0$
C
With time period $\frac{2\pi}{\omega}$.
✓
With time period $\frac{\pi}{\omega}.$

Answer

Correct option: D.

With time period $\frac{\pi}{\omega}.$

Given equation:
$\text{x}=\text{x}_0\sin^2\omega\text{t}$
$\Rightarrow\text{x}=\frac{\text{x}_0}{2}\big(\cos2\omega\text{t}-1\big)$
Now, the amplitude of the particle is $=\frac{\text{x}_0}{2}$ and the angular frequency of the $\text{SHM}$ is $2\omega.$
Thus, time period of the $\text{SHM} = \frac{2\pi}{\text{angular frequecy}}=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}$

View full question & answer→

MCQ 221 Mark

In a simple harmonic motion:

A
The potential energy is always equal to the kinetic energy.
B
The potential energy is never equal to the kinetic energy.
C
The average potential energy in any time interval is equal to the average kinetic energy in that time interval.
✓
The average potential energy in one time period is equal to the average kinetic energy in this period.

Answer

Correct option: D.

The average potential energy in one time period is equal to the average kinetic energy in this period.

The kinetic energy of the motion is given as,
$\frac{1}{2}\text{K}\text{A}^2\cos^2\omega\text{t}$
The potential energy is calculated as,
$\frac{1}{2}\text{K}\text{A}^2\sin^2\omega\text{t}$
As the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.

View full question & answer→

MCQ 231 Mark

A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken to the poles it will:

A
Run slow.
B
Run fast.
C
Stop working.
✓
Give correct time.

Answer

Correct option: D.

Give correct time.

Because the time period of a spring-mass system does not depend on the acceleration due to gravity.

View full question & answer→

MCQ 241 Mark

Figure represents two simple harmonic motions. The parameter which has different values in the two motions is:

A
Amplitude.
B
Frequency.
✓
Phase.
D
Maximum velocity.

Answer

Correct option: C.

Phase.

Because the direction of motion of particles A and B is just opposite to each other.

View full question & answer→

MCQ 251 Mark

Which of the following quantities are always positive in a simple harmonic motion?

✓
$\overrightarrow{\text{F}}.\overrightarrow{\text{a}}$
B
$\overrightarrow{\text{v}}.\overrightarrow{\text{r}}$
C
$\overrightarrow{\text{a}}.\overrightarrow{\text{r}}$
D
$\overrightarrow{\text{F}}.\overrightarrow{\text{r}}$

Answer

Correct option: A.

$\overrightarrow{\text{F}}.\overrightarrow{\text{a}}$

As the direction of force and acceleration are always same, $\overrightarrow{\text{F}}.\overrightarrow{\text{a}}.$ is always positive.

View full question & answer→

MCQ 261 Mark

The displacement of a particle in simple harmonic motion in one time period is:

A
A
B
2A
C
4A
✓
zero.

Answer

Correct option: D.

zero.

Displacement is defined as the distance between the starting and the end point through a straight line. In one complete oscillation, the net displacement is zero as the particle returns to its initial position.

View full question & answer→

MCQ 271 Mark

The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point, the box is released to fall freely. As seen from the box during this period, the bob will:

A
Continue its oscillation as before.
B
Stop.
✓
Will go in a circular path.
D
Move on a straight line.

Answer

Correct option: C.

Will go in a circular path.

As the acceleration due to gravity acting on the bob of pendulum, due to free fall gives a torque to the pendulum, the bob goes in a circular path.

View full question & answer→

MCQ 281 Mark

The motion of a particle is given by $\text{x = A}\sin\omega\text{t+B}\cos\omega\text{t}.$ The motion of the particle is:

A
Not simple harmonic.
B
Simple harmonic with amplitude A + B.
C
Simple harmonic with amplitude $\frac{\text{(A}+\text{B)}}{2}.$
✓
Simple harmonic with amplitude $\sqrt{(\text{A}^2+\text{B}^2)}.$

Answer

Correct option: D.

Simple harmonic with amplitude $\sqrt{(\text{A}^2+\text{B}^2)}.$

$\text{x}=\text{A}\sin\omega\text{t}+\text{B}\cos\omega\text{t}\ ...(1)$
Acceleration,
$\text{a}=\frac{\text{d}^2\text{x}}{\text{d}\text{t}^2}=\frac{\text{d}^2}{\text{dt}^2}(\text{A}\sin\omega\text{t}+\text{B}\cos\omega\text{t})$
$=\frac{\text{d}}{\text{dt}}(\text{A}\omega\cos\omega\text{t}-\text{B}\omega\sin\omega\text{t)}$
$=-\text{A}\omega^2\sin\omega\text{t}=\text{B}\omega^2\cos\omega\text{t}$
$=-\omega^2(\text{A}\sin\omega\text{t}+\text{B}\cos\omega\text{t})$
$=-\omega^2\text{x}$
For a body to undergo simple harmonic motion,
acceleration, a = -kx .....(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude, $\text{A}=\sqrt{\text{A}^2+\text{B}^2}.$

View full question & answer→

Generate Paper Free All Simple Harmonic Motion topics