A steel strip, clamped at one end, vibrates with a frequency of 20 Hz and a amplitude of 5 mm at the free end, where a small mass of 2 g is positioned. Find

(i) the velocity of the end when passing through the zero position,
(ii) the acceleration at maximum displacement,
(iii) the maximum kinetic energy of the mass.

Please help me and solve

Simple harmonic motion

please I need both the three anwers

1),2),3).

please help me solve

Answer please

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To solve this problem, we can use the principles of simple harmonic motion. Here's how you can find each of the required quantities:

(i) The velocity of the end when passing through the zero position:
In simple harmonic motion, the velocity of an object is zero at the extreme positions (maximum displacement) and reaches its maximum value as it passes through the equilibrium (zero) position. The relationship between velocity and displacement in simple harmonic motion is given by the equation:

v = ω * A * cos(ωt)

Where:
v = velocity of the object
ω = angular frequency (2πf, where f is the frequency)
A = amplitude of motion (maximum displacement)
t = time

Since the velocity is zero at the extreme positions, we need to find the velocity at the equilibrium position. In this case, the amplitude is given as 5 mm, and the frequency is 20 Hz. First, convert the amplitude to meters:

A = 5 mm = 0.005 m

Next, calculate the angular frequency:

ω = 2πf
ω = 2π * 20 Hz
ω = 40π rad/s

Substituting these values into the equation:

v = (40π rad/s) * (0.005 m) * cos(40πt)
v = 0.2π cos(40πt) (approx.)

So, the velocity of the end when passing through the zero position is 0.2π m/s.

(ii) The acceleration at maximum displacement:
The acceleration of an object in simple harmonic motion is directly proportional to its displacement and is given by the equation:

a = -ω^2 * x

Where:
a = acceleration of the object
ω = angular frequency
x = displacement from the equilibrium position

At maximum displacement (A), the acceleration is a maximum as well. The angular frequency is already calculated as 40π rad/s. Substituting these values into the equation:

a = -(40π)^2 * 0.005 m
a = -8000π^2 m/s^2 (approx.)

So, the acceleration at maximum displacement is -8000π^2 m/s^2.

(iii) The maximum kinetic energy of the mass:
The maximum kinetic energy of a mass in simple harmonic motion is given by the equation:

K.E. = (1/2) * m * ω^2 * A^2

Where:
K.E. = maximum kinetic energy
m = mass of the object
ω = angular frequency
A = amplitude of motion

The mass is given as 2 g, which can be converted to kg:

m = 2 g = 0.002 kg

Substituting the given values:

K.E. = (1/2) * (0.002 kg) * (40π rad/s)^2 * (0.005 m)^2
K.E. = 0.0008π^2 J (approx.)

So, the maximum kinetic energy of the mass is approximately 0.0008π^2 Joules.

1). v= 0.