Equation of SHM - NEET Physics Questions
Question 41: easy

Assertion (A): \(x = A sin \omega t\) & \(y = B cos \omega t\) In the above co-ordinates particle moves in elliptical path.


Reason (R): A periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitude.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: From \(x = A sin \omega t\) and \(y = B cos \omega t\), we get \(\left( \frac{x}{A} \right)^2 + \left( \frac{y}{B} \right)^2 = 1\), which is an ellipse. Reason (R) is true: Fourier's theorem states any periodic motion can be decomposed into harmonic components. However, (R) does not explain why the given equations describe an ellipse.

Question 42: easy

Assertion (A): When a simple pendulum is made to oscillate on the surface of moon, its time period increases.


Reason (R): Gravity at moon is less than gravity at earth.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: The period of a simple pendulum is inversely proportional to the square root of 'g' (\(T = 2pi sqrt{l/g}\)). Reason (R) is true: Gravity on the moon is significantly less than on Earth. Since 'g' decreases, the period 'T' increases, making (R) the correct explanation for (A).

Question 43: easy

Assertion (A): \(x = \sin^2(\omega t)\) represents a SHM about mean position \(x = \frac{1}{2}\).


Reason (R): \(a \propto -x\) is the necessary condition for SHM.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A): \(x = \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}\). Let \(y = x - \frac{1}{2} = -\frac{1}{2}\cos(2\omega t)\). This is SHM about \(x = \frac{1}{2}\). So (A) is true. Reason (R): For SHM, acceleration is proportional to negative displacement \(a = -\omega^2 x\). So (R) is true. However, (R) does not explain (A).

Question 44: easy

Assertion (A): If PE of a particle executing SHM is given by \(U = x^2 – 10x + 27\), then it is executing SHM about \(x = 5\).


Reason (R): At mean position, restoring force is zero.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A): Given \(U = x^2 - 10x + 27 = (x-5)^2 + 2\). For SHM, \(U = \frac{1}{2}k(x-x_0)^2 + U_0\). Comparing, mean position \(x_0 = 5\). So (A) is true. Reason (R): The restoring force \(F = -\frac{dU}{dx}\). At equilibrium (mean) position, \(F=0\). So (R) is true. (R) does not explain (A).

Question 45: easy

Assertion (A): General vibrations of a polyatomic molecule about its equilibrium position is periodic but not SHM.


Reason (R): A periodic motion can always be expressed as a sum of infinite number of harmonic motion with appropriated amplitude.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true. Complex vibrations of polyatomic molecules are periodic but generally not simple harmonic motion (SHM). Reason (R) is true. This is the principle of Fourier analysis, stating that any periodic motion can be decomposed into a sum of simple harmonic components. Reason (R) correctly explains why complex periodic motions (like polyatomic vibrations) are not SHM but can still be described as periodic.

Question 46: easy

Assertion (A): In SHM acceleration leads displacement by phase \(\pi\).


Reason (R): In SHM velocity leads displacement by phase \(\pi/2\).


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true. If displacement \(x = Asin(\omega t)\), then acceleration \(a = -A\omega^2sin(\omega t) = A\omega^2sin(\omega t + \pi)\). Reason (R) is true. Velocity \(v = A\omega cos(\omega t) = A\omega sin(\omega t + \pi/2)\). Both statements are true, but the phase relationship of velocity with displacement does not explain the phase relationship of acceleration with displacement directly; they are separate facts of SHM.

Question 47: easy

Assertion (A): Amplitude of SHM \(x = 4sin^2\omega t + 2cos^2\omega t + 2sin\omega t cos\omega t\) is \(sqrt{2}\).


Reason (R): Angular frequency of given equation is \(2\omega\).


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

The expression \(x = 4sin^2\omega t + 2cos^2\omega t + 2sin\omega t cos\omega t\) simplifies to \(x = 3 + sin(2\omega t) - cos(2\omega t)\). The oscillatory part is \(sin(2\omega t) - cos(2\omega t)\). Assertion (A) is true, its amplitude is \(\sqrt{1^2 + (-1)^2} = \sqrt{2}\). Reason (R) is true, the angular frequency is \(2\omega\). However, the angular frequency does not explain the specific amplitude value.

Question 48: easy

Assertion (A): The graph between velocity and displacement for a harmonic oscillator is a parabola.


Reason (R): Velocity does change uniformly with displacement in harmonic motion.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

For a harmonic oscillator, velocity \( v \) and displacement \( x \) are related by \( v = \omega \sqrt{A^2 - x^2} \). Squaring this gives \( v^2 = \omega^2 (A^2 - x^2) \), which is an equation of an ellipse, not a parabola. So (A) is false. Velocity does not change uniformly with displacement, hence (R) is also false. Thus, both A and R are false.

Question 49: easy

Assertion (A): Sine and cosine functions are periodic functions.


Reason (R): Sinusoidal functions repeat its values after a definite interval of time.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Periodic functions like sine and cosine repeat their values over a fixed period. Reason (R) defines periodicity, which directly explains Assertion (A).
Thus, both are true, and R explains A.

Question 50: easy

Assertion (A): In SHM the velocity is maximum when the acceleration is minimum.


Reason (R): Displacement and velocity in SHM differ in phase by \(\frac{\pi}{2}\) .


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

In SHM, velocity is max at equilibrium (where displacement is zero), and acceleration is min (zero) at equilibrium. So A is true.
Displacement `\(x = A\sin(\omega t)\)` and velocity `\(v = A\omega\cos(\omega t)\)` differ in phase by \(\frac{\pi}{2}\). So R is true.
However, R explains phase relation, not why maximum velocity occurs at minimum acceleration. Hence, R does not explain A.