Physics M.C.Q (1 Marks)

4. $\sqrt{6}$ times slower.

Explanation:

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:

$\text{T}_{\text{moon}}=2\pi\sqrt{\frac{\text{l}}{\text{g}_{\text{moon}}}}=2\pi\sqrt{\frac{\text{l}}{\Big(\frac{\text{g}}{6}\Big)}}$

$=\sqrt{6}\Big(2\pi\sqrt{\frac{\text{l}}{\text{g}}}\Big)=\sqrt{6}\text{T}$

1. The maximum potential energy equals the maximum kinetic energy.
2. The minimum potential energy equals the minimum kinetic energy.

Explanation:

In SHM,

maximum kinetic energy $=\frac{1}{2}\text{kA}^2$

maximum potential energy $=\frac{1}{2}\text{kA}^2$

The minimum value of both kinetic and potential energy is zero. 

Therefore, in a simple harmonic motion the maximum kinetic energy and maximum potential energy are equal. Also, the minimum kinetic energy and the minimum potential energy are equal.

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

Explanation:

$\text{x}=\text{A}\sin\omega\text{t}+\text{B}\cos\omega\text{t}\ ...(1)$

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, $\text{A}=\sqrt{\text{A}^2+\text{B}^2}.$

1. on a straight line.

3. Periodic.

4. Simple harmonic.

Explanation:

The given equation is a solution to the equation of simple harmonic motion. The amplitude is $\big(\overrightarrow{\text{i}}+2\overrightarrow{\text{j}}\big)\text{A},$ following equation of straight line y = mx + c. Also, a simple harmonic motion is periodic.

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

Explanation:

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 k₁ is given as,

$\text{k}_1=2\text{k}$

New time period T₁:

$\text{T}_1=2\pi\sqrt{\frac{\text{m}}{2\text{k}}}=\frac{1}{\sqrt{2}}2\pi\sqrt{\frac{\text{m}}{\text{k}}}=\frac{1}{\sqrt{2}}\text{T}$ 

3. $\overrightarrow{\text{a}}.\overrightarrow{\text{r}}$

4. $\overrightarrow{\text{F}}.\overrightarrow{\text{r}}$

Explanation:

In S.H.M.,

F = -kx

Therefore, $\overrightarrow{\text{F}}.\overrightarrow{\text{r}}$ will always be negative. As acceleration has the same direction as the force, $\overrightarrow{\text{a}}.\overrightarrow{\text{r}}$ will also be negative, always.

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

Explanation:

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

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

Explanation:

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,

$\text{A}\omega_\text{A}=\text{B}\omega_\text{B}$

$\Rightarrow\text{A}\sqrt{\frac{\text{k}_1}{\text{m}_\text{A}}}=\text{B}\sqrt{\frac{\text{k}_2}{\text{m}_\text{B}}}$

$\Rightarrow\text{A}\sqrt{\frac{\text{k}_1}{\text{m}}}=\text{B}\sqrt{\frac{\text{k}_2}{\text{m}}}\big(\text{m}_\text{A}=\text{m}_{\text{B}}=\text{m}\big)$

$\Rightarrow\frac{\text{A}}{\text{B}}=\sqrt{\frac{\text{k}_2}{\text{k}_1}}$

4. Remain E.

Explanation:

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 m₁ 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 E₁ is given as,

$\text{E}_1=\frac{1}{2}\text{m}_1\omega_1^2\text{A}^2$

$=\frac{1}{2}\big(2\text{m}\big)\Big(\sqrt{\frac{\text{k}}{2\text{m}}}\Big)^2\text{A}^2$

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

4. $\text{Zero}.$

Explanation:

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,

$\text{A}\omega^2+(-\text{A}\omega^2)=0$

1. As x increases k increases.

Explanation:

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.

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

2. $\overrightarrow{\text{v}}\times\overrightarrow{\text{r}}$

3. $\overrightarrow{\text{a}}\times\overrightarrow{\text{r}}$

4. $\overrightarrow{\text{F}}\times\overrightarrow{\text{r}}$

Explanation:

As $\overrightarrow{\text{F}},\overrightarrow{\text{a}},\overrightarrow{\text{r}}$ and $\overrightarrow{\text{v}}$ are either parallel or anti-parallel to each other, their cross products will always be zero.

2. B

Explanation:

 At t = 0,

 Displacement (x₀) 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 - x₀ = A + B - A = B

3. On a circle.

Explanation:

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^o 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.

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

Explanation:

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 SHM is $2\omega.$

Thus, time period of the SHM $\frac{2\pi}{\text{angular frequecy}}=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}$

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

Explanation:

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

3. Its length should be decreased to keep correct time.

Explanation:

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.

3. Remain same.

Explanation:

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.

1. Displacement from the mean position.

Explanation:

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.

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

Explanation:

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.