Simple Pendulum - Derivation of Expression for its Time Period NEET Questions

Physics - Oscillations and Waves

Simple Pendulum - Derivation of Expression for its Time Period MCQ Questions


7.
By Newton's law of rotational motion, τ = Iα. For a simple pendulum where the string is massless, the moment of inertia I about the support is:
A.
I = (1/3)mL²
B.
I = m/L²
C.
I = mL²
D.
I = mL
😑 View Answer Rough Work Error
ANSWER :
C. I = mL²
8.
Substituting τ = Iα into the torque equation for the pendulum gives:
A.
mL²α = mgL sinθ, so α = (g/L) sinθ
B.
mL²α = −mgL tanθ
C.
mL²α = −mgL sinθ, so α = −(g/L) sinθ
D.
mLα = −mg sinθ, so α = −(g/L) sinθ
😑 View Answer Rough Work Error
ANSWER :
C. mL²α = −mgL sinθ, so α = −(g/L) sinθ
9.
The small-angle approximation used in the pendulum derivation is:
A.
cosθ ≈ 1 for small θ
B.
tanθ ≈ θ for small θ
C.
sinθ ≈ θ (in radians) for small θ
D.
sinθ ≈ θ² for small θ
😑 View Answer Rough Work Error
ANSWER :
C. sinθ ≈ θ (in radians) for small θ
10.
For what maximum angle is the small-angle approximation (sinθ ≈ θ) reasonably accurate (error < 2%)?
A.
Up to about 20°
B.
Up to about 45°
C.
Up to about 60°
D.
Up to about 5°
😑 View Answer Rough Work Error
ANSWER :
A. Up to about 20°
11.
After applying sinθ ≈ θ, the equation of motion α = −(g/L)sinθ becomes:
A.
α = (g/L)θ, which is exponential growth
B.
α = −(g²/L)θ
C.
α = −(g/L)θ, which is the SHM condition
D.
α = −(g/L)θ², which is non-linear SHM
😑 View Answer Rough Work Error
ANSWER :
C. α = −(g/L)θ, which is the SHM condition
12.
Comparing α = −(g/L)θ with the standard SHM form α = −ω²θ, the angular frequency ω of the pendulum is:
A.
ω = L/g
B.
ω = g/L
C.
ω = √(L/g)
D.
ω = √(g/L)
😑 View Answer Rough Work Error
ANSWER :
D. ω = √(g/L)