Question

A block of mass m = 1kg is placed on a smooth surface and is connected with a spring of constant K = 100N/m and another end of the spring is connected to a fixed wall as shown. The block is pulled by a distance A = 0.10 m from its natural length and released at t = 0

m =1.0kg k = 100N/m 5 Smooth 01m

• A. The maximum speed is after $t=\frac{\pi }{20}s$
• B. Time taken to cover the first 0.10 m, t=π/20s
• C. Time taken to cover the first 0.05 m, t=π/40s
• D. Time taken to cover the first 0.05 m, t=π/30s

Answer 1

The correct option is A

The maximum speed is after $t=\frac{\pi }{20}s$

Step1: Given data

$k=100N/m,m=1kg,L=0.1m$

Step2: Formula used

The formula for the angular frequency of spring in simple harmonic motion is $\omega =\sqrt{\frac{K}{m}}[whereK=cons\tan t,m=mass]$

$T=\frac{2\pi }{\omega }[whereT=time,\omega =angularfrequency]$

Step3: Calculating angular acceleration

$$\begin{gathered} \omega =\sqrt{\frac{K}{m}} \\ \omega =\sqrt{\frac{100}{1}} \\ \omega ==10rad/s \end{gathered}$$

Step4: Calculating time taken

The maximum speed of the spring will be at its natural length,

$$\begin{gathered} \frac{T}{4}=\frac{2\pi }{10\times 4} \\ \frac{T}{4}=\frac{\pi }{20}s \end{gathered}$$

Now, the time taken to cover $0.1m$ is $\frac{T}{4}=\frac{\pi }{20}s$

Now, the time taken to cover 0.05m will be $\frac{T}{4}\times \frac{2}{3}=\frac{\pi }{20}\times \frac{2}{3}=\frac{\pi }{30}s$

Therefore, the maximum speed is after $t=\frac{\pi }{20}s$.

Hence, option A is the correct answer.