Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.

Frequency f is the reciprocal of the period T. Thus, f = 1/T = 1/4 = 0.25 Hz.

Maximum speed (v_max) = ωA. Thus, ω = v_max/A = 4/2 = 2 rad/s.

The period of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T ≈ 1.41 s.

The period T of a simple pendulum is given by T = 2π√(L/g). Here, L = 1 m and g = 9.8 m/s². Thus, T = 2π√(1/9.8) ≈ 2.0 s.

The period T of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T = 1.41 s.

The period of a pendulum is given by T = 2π√(L/g). If L is increased to 4L, T becomes 2π√(4L/g) = 2T = 2 seconds.

The period T = 2π√(L/g). If L is increased by a factor of 4, T increases by a factor of √4 = 2.

The period T = 2π√(L/g). If L is tripled, T becomes √3 times longer, so T = 2 s.

Length L = (gT^2)/(4π^2) = (9.8 * 1^2)/(4 * π^2) ≈ 1 m.

Angular frequency (ω) = 2π/T = 2π/1.5 = 4π/3 rad/s.

Frequency (f) is the reciprocal of the period (T). Therefore, f = 1/T = 1/2 = 0.5 Hz.

The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T²g)/(4π²). Using g = 9.8 m/s², L = (2² * 9.8)/(4π²) ≈ 1 m.

The period T of a simple pendulum is given by T = 2π√(L/g). For length 4L, T' = 2π√(4L/g) = 2T.

In simple harmonic motion, the restoring force is proportional to the displacement from the equilibrium position.

A pendulum swinging with a small amplitude exhibits simple harmonic motion.

The time period (T) is the reciprocal of frequency (f). T = 1/f = 1/5 = 0.2 s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 0.5 kg. Thus, ω = √(200/0.5) = √400 = 20 rad/s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 2 kg. Thus, ω = √(200/2) = √100 = 10 rad/s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 50 N/m and m = 2 kg. Thus, ω = √(50/2) = √25 = 5 rad/s.

ω = V_max/A = 2/0.1 = 20 rad/s.

Maximum potential energy (PE) = (1/2)kA^2 = (1/2)(200)(0.1^2) = 1 J.

T = 2π√(m/k) = 2π√(2/200) = 1 s.

The total energy E is proportional to the square of the amplitude. If the amplitude is halved, the new energy will be E/4.

Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.

Total energy in SHM is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 4.

The maximum speed v_max of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.

The maximum speed (v_max) of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.

The maximum displacement in simple harmonic motion is equal to the amplitude, which is 5 cm.

Using the wave equation v = fλ, we can find λ = v/f = 340/170 = 2 m.